RD Sharma Class 12 Ex 30.2 Solutions Chapter 30 Linear Programming

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter30
Exercise30.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 30.2 Solutions Chapter 30 Linear Programming

Question 1. Maximize Z = 5x + 2y, 

Subject to

3x + 5y ≤ 15

5x + 2y ≤ 10

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

3x + 5y = 15, 

5x + 2y = 10, 

x = 0 and 

y = 0

Area represented by 3x + 5y ≤ 15:

The line 3x + 5y = 15 connects the coordinate axes at A(5,0) and B(0,3) respectively. 

By connecting these points we will get the line 3x + 5y = 15.

Thus,

(0,0) assure the in equation 3x + 5y ≤ 15. 

Hence,

The area having the origin shows the solution set of the in equation 3x + 5y ≤ 15.

Area shows by 5x + 2y ≤ 10:

The line 5x + 2y = 10 connects the coordinate axes at C(2,0) and D(0,5) respectively. 

By connecting these points we will get the line 5x + 2y = 10.

Thus, 

(0,0) satisfies the in equation 5x + 2y ≤ 10. 

Hence,

The area having the origin shows the solution set of the in equation 5x + 2y ≤ 10.

Area shows by x ≥ 0 and y ≥ 0:

Here, 

All the point in the first quadrant assures these in equations. 

Thus, 

The first quadrant is the area shows by the in equations x ≥ 0, and y ≥ 0.

The feasible area determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

 The corner points of the feasible area are O(0, 0), C(2, 0), E\left(\frac{20}{19},\ \frac{45}{19}\right)    and B(0, 3).

The values of Z at these corner points are as follows.

\left(\frac{20}{19},\ \frac{45}{19}\right)

Hence, 

The maximum value of Z 

\frac{235}{19} at the point \left(\frac{20}{19},\ \frac{45}{19}\right)   

Hence, 

x= \frac{20}{19} and y = \frac{45}{19}   is the best solution of the given LPP.

Therefore, 

The best value of Z is \frac{235}{19}.

Question 2. Maximize Z = 9x + 3y

Subject to

2x + 3y ≤ 13

3x + y ≤ 5

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

2x + 3y = 13, 

3x +y = 5, 

x = 0 and 

y = 0

Area shown by 2x + 3y ≤ 13:

The line 2x + 3y = 13 connects the coordinate axes at A \left(\frac{13}{2},\ 0\right) and B \left(0,\ \frac{13}{3}\right) respectively.

By connecting these points we will get the line 2x + 3y = 13.

Thus,

(0,0) assures the line equation 2x + 3y ≤ 13. 

Thus, the area showing the origin represents the solution set of the in equation 2x + 3y ≤ 13.

The area shows by 3x + y ≤ 5:

The line 5x + 2y = 10 connects the coordinate axes at C \left(\frac{5}{3},\ 0\right)   and D(0, 5) respectively. 

After connecting these points, 

We will get the line 3x + y = 5.

Thus,

(0,0) assures the in equation 3x + y ≤ 5. 

Thus,

The area having the origin represents the solution set of the in equation 3x + y ≤ 5.

Area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assures these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0, and y ≥ 0.
 

The suitable area determined by the system of constraints, 

2x + 3y ≤ 13, 

3x + y ≤ 5, 

x ≥ 0, and 

y ≥ 0, are as follows.

The corner points of the suitable area are O(0, 0), C \left(\frac{5}{3},\ 0\right), E \left(\frac{2}{7},\ \frac{29}{7}\right) and B \left(0,\ \frac{13}{3}\right).

The values of Z at these corner points are as follows.

\left(\frac{5}{3},\ 0\right)

Here,

We can see that the maximum value of the objective function Z is 15 which is at C \left(\frac{5}{3},\ 0\right) and E \left(\frac{2}{7},\ \frac{29}{7}\right).

Thus, 

The best value of Z is 15.

Question 3. Minimize Z = 18x + 10y

Subject to

4x + y ≥ 20

2x + 3y ≥ 30

x, y ≥ 0

Solution:

Convert the given in equations into equations,

We will get the following equations:

4x + y = 20, 

2x +3y = 30, 

x = 0 and 

y = 0

Area shown by 4x + y ≥ 20 :

The line 4x + y = 20 connects the coordinate axes at A(5, 0) and B(0, 20) respectively. 

By connecting these points we will get the line 4x + y = 20.

Thus,

(0,0) does not assure the in equation 4x + y ≥ 20. 

Thus,

The area in xy plane which does not have the origin represents the solution set of the in equation 4x + y ≥ 20.

Area shown by 2x +3y ≥ 30:

The line 2x +3y = 30 connects the coordinate axes at C(15,0) and D(0, 10) respectively. 

By joining these points we will get the line 2x +3y = 30.

Thus, (0,0) does not assure the in equation 2x +3y ≥ 30.

Thus,

The area which does not have the origin shows the solution set of the in equation 2x +3y ≥ 30.

Area shown by x ≥ 0 and y ≥ 0:

Hence, 

Every point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints, 4x + y ≥ 20, 2x +3y ≥ 30, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the suitable area are 

B(0, 20), 

C(15,0), 

E(3,8) and 

C(15,0).

The values of Z at these corner points are as follows.

Corner pointZ = 18+ 10y
B(0, 20)18 × 0 + 10 × 20 = 200
E(3,8)18 × 3 + 10 × 8 = 134
C(15,0) 18 × 15 + 10 ×0 = 270

Hence, 

The minimum value of Z is 134 at the point E(3,8). Hence, x = 3 and y =8 is the best solution of the given LPP.

Therefore, 

The best value of Z is 134.

Question 4.  Maximize Z = 50x + 30y

Subject to

2x + y ≤ 18

3x + 2y ≤ 34

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

2x + y = 18, 3x + 2y = 34

The area shown by 2x + y ≥ 18:

The line 2x + y = 18 connects the coordinate axes at A(9, 0) and B(0, 18) respectively. 

After connecting these points 

We will get the line 2x + y = 18.

Thus,

(0,0) does not assure the in equation 2x + y ≥ 18. 

Thus,

The region in xy plane which does not have the origin represents the solution set of the in equation 2x + y ≥ 18.

The area shown by 3x + 2y ≤ 34:

The line 3x + 2y = 34 connects the coordinate axes at C \left(\frac{34}{3},\ 0\right), 0 and D(0, 17) respectively. 

After joining these points we will get the line 3x + 2y = 34.

Thus,

(0,0) assure the in equation 3x + 2y ≤ 34. 

Thus,

The area having the origin shows the solution set of the in equation 3x + 2y ≤ 34.

The corner points of the suitable area are 

A(9, 0), C \left(\frac{34}{3},\ 0\right)   and E(2, 14).

The values of Z at these corner points are as follows.

\left(\frac{34}{3},\ 0\right)

Hence, 

The maximum value of Z is \frac{1700}{3} at the point \left(\frac{34}{3},\ 0\right).

Hence,

x = \frac{34}{3} and y = 0 is the best solution of the given LPP.

Therefore,

The best value of Z is \frac{1700}{3}

Question 5. Maximize Z = 4x + 3y

Subject to

3x + 4y ≤ 24

8x + 6y ≤ 48

x ≤ 5

y ≤ 6

x, y ≥ 0

Solution:

Here we have to maximize Z = 4x + 3y

Convert the given in equations into equations, 

We will get the following equations:

3x + 4y = 24, 

8x + 6y = 48, 

x = 5, 

y = 6, 

x = 0 and 

y = 0.

The line 3x + 4y = 24 connects the coordinate axis at A(8, 0) and B(0,6). 

Connect these points to get the line 3x + 4y = 24.

Thus, 

(0, 0) assure the in equation 3x + 4y ≤ 24.

Thus, 

The area in xy-plane that have the origin showing the solution set of the given equation.

The line 8x + 6y = 48 connects the coordinate axis at C(6, 0) and D(0,8). Connect these points to get the line 8x + 6y = 48.

Thus,

(0, 0) assures the in equation 8x + 6y  ≤ 48. 

Thus, 

The area in xy-plane that have the origin shows the solution set of the given equation.

x = 5 is the line passing through x = 5 parallel to the Y axis.

y = 6 is the line passing through y = 6 parallel to the X axis.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shows by the in equations.

These lines are drawn using a suitable scale.

The corner points of the suitable area are O(0, 0), G(5, 0), F \left(5,\ \frac{4}{3}\right)  , E \left(\frac{24}{7},\ \frac{24}{7}\right)   

and B(0, 6).

The values of Z at these corner points are as follows.
 

\left(5,\ \frac{4}{3}\right)

Here we can see that the maximum value of the objective function Z is 24 

Which is at F \left(5,\ \frac{4}{3}\right)   and E 

Therefore,

The best value of Z is 24.

Question 6. Maximize Z = 15x + 10y

Subject to

3x + 2y ≤ 80

2x + 3y ≤ 70

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

3x + 2y = 80, 2x + 3y = 70, x = 0 and y = 0

The area shown by 3x + 2y ≤ 80 :

The line 3x + 2y = 80 connects the coordinate axes at A \left(\frac{80}{3},\ 0\right)  , 0 and B(0, 40) respectively. 

By connecting these points we will get the line 3x + 2y = 80.

Thus,

(0,0) assure the in equation 3x + 2y ≤ 80 . 

Thus,

The area having the origin represents the solution set of the in equation 3x + 2y ≤ 80 .

The area shown by 2x + 3y ≤ 70:

The line 2x + 3y = 70 connects the coordinate axes at C(35, 0)C35, 0 and D(0, 703)D0, 703 respectively. 

After connecting these points we will get the line 2x + 3y ≤ 70.

Thus,

(0,0) assure the in equation 2x + 3y ≤ 70. 

Thus,

The area having the origin shows the solution set of the in equation 2x + 3y ≤ 70.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

Every point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0 and y ≥ 0.
 

The suitable area determined by the system of constraints, 3x + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the suitable area are O(0, 0), A \left(\frac{80}{3},\ 0\right)  , 0 ,E(20, 10) and D \left(0,\ \frac{70}{3}\right)  .

The values of Z at these corner points are as follows.

\left(\frac{80}{3},\ 0\right)

Thus we can see that the maximum value of the objective function Z is 400 which is at A \left(\frac{80}{3},\ 0\right)

and E(20, 10).

Therefore, 

The best value of Z is 400.

Question 7.  Maximize Z = 10x + 6y

Subject to

3x + y ≤ 12

2x + 5y ≤ 34

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

3x + y = 12, 2x + 5y = 34, x = 0 and y = 0

The area shown by 3x + y ≤ 12:

The line 3x + y = 12 connects the coordinate axes at A(4, 0) and B(0, 12) respectively. 

After connecting these points we will get the line 3x + y = 12.

Thus,

(0,0) assure the in equation 3x + y ≤ 12. 

Thus,

The area having the origin represents the solution set of the in equation 3x + y ≤ 12.

The area shown by 2x + 5y  ≤ 34:

The line 2x + 5y = 34 connects the coordinate axes at C(17, 0) and D (0,\ \frac{34}{5})   respectively. 

After connecting these points we will get the line 2x + 5y ≤ 34.

Thus,

(0,0) assure the in equation 2x + 5y ≤ 34. 

Thus,

The area having the origin represents the solution set of the in equation 2x + 5y ≤ 34.

Area having by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0 and y ≥ 0.
 

The suitable area determined by the system of constraints, 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the suitable area are O(0, 0), A(4, 0) ,E(2, 6) and D(0,\ \frac{34}{5})  .

The values of Z at these corner points are as follows:

(0,\ \frac{34}{5})

Here we can see that the maximum value of the objective function Z is 56 which is at E(2, 6) that means at x = 2 and y = 6.

Therefore, 

The best value of Z is 56.

Question 8.  Maximize Z = 3x + 4y

Subject to

2x + 2y ≤ 80

2x + 4y ≤ 120

Solution:

Here we have to maximize Z = 3x + 4y

Convert the given in equations into equations, we will get the following equations:

2x + 2y = 80, 2x + 4y = 120

The area shown by 2x + 2y ≤ 80:

The line 2x + 2y = 80 connects the coordinate axes at A(40, 0) and B(0, 40) respectively. 

After connecting these points we will get the line 2x + 2y = 80.

Thus, 

(0,0) assure the in equation 2x + 2y ≤ 80. 

Thus,

The area having the origin represents the solution set of the in equation 2x + 2y ≤ 80.

The area shown by 2x + 4y ≤ 120:

The line 2x + 4y = 120 connects the coordinate axes at C(60, 0) and D(0, 30) respectively. 

After connecting these points we will get the line 2x + 4y ≤ 120.

Thus,

(0,0) assure the in equation 2x + 4y ≤ 120. 

Thus,

The area having the origin represents the solution set of the in equation 2x + 4y ≤ 120.

The suitable area determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:

The corner points of the suitable area are O(0, 0), A(40, 0), E(20, 20) and D(0, 30).

The values of Z at these corner points are as follows:

Corner pointZ = 3+ 4y
O(0, 0)3 × 0 + 4 × 0 = 0
A(40, 0)3× 40 + 4 × 0 = 120
E(20, 20)3 × 20 + 4 × 20 = 140
D(0, 30) 10 × 0 + 4 ×30 = 120

Here we can see that the maximum value of the objective function Z is 140 which is at E(20, 20) that means at

x = 20 and y = 20.

Therefore, 

The best value of Z is 140.

Question 9. Maximize Z = 7x + 10y

Subject to

x + y ≤ 30000

       y ≤ 12000

       x ≥ 6000

       x ≥ y

   x, y ≥ 0

Solution:

Here we have to maximize Z = 7x + 10y

Convert the given in equations into equations, 

We will get the following equations:

x + y = 30000,y = 12000, x = 6000, x = y, x = 0 and y = 0.

Region represented by x + y ≤ 30000:

The line x + y = 30000 connects the coordinate axes at A(30000, 0) and B(0, 30000) respectively. 

After connecting these points we will get the line x + y = 30000.

Thus, (0,0) assure the in equation x + y ≤ 30000. 

Thus,

the area having the origin represents the solution set of the in equation x + y ≤ 30000.

The line y = 12000 is the line that passes through C(0,12000) and parallel to x axis.

The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.

The area shown by x ≥ y

The line x = y is the line that passes through origin.The points to the right of the line x = y assure the inequation x ≥ y.

Like by taking the point (−12000, 6000).

Here, 6000 > −12000 which implies y > x. 

Hence, 

the points to the left of the line x = y will not assure the given inequation x ≥ y.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assure these inequations. 

Thus, 

the first quadrant is the area shown by the inequations x ≥ 0 and y ≥ 0.

The suitable area determined by the system of constraints, x + y ≤ 30000, y ≤ 12000, x ≥ 6000, x ≥  y , x ≥ 0 and y ≥ 0 are as follows:

The corner points of the feasible region are 

D(6000, 0), 

A(3000, 

F(18000, 12000) and 

E(12000, 12000).

The values of Z at these corner points are as follows:

Corner pointZ = 7+ 10y
D(6000, 0)7 × 6000 + 10 × 0 = 42000
A(3000, 0)A3000, 07× 3000 + 10 × 0 = 21000
F(18000, 12000)7 × 18000 + 10 × 12000 = 246000
E(12000, 12000) 7 × 12000 + 10 ×12000 = 204000

Here we can see that the maximum value of the objective function Z is 246000 which is at F(18000, 12000) that means at

x = 18000 and y = 12000.

Therefore, 

The best value of Z is 246000.

Question 10. Minimize Z = 2x + 4y

Subject to

x + y ≥ 8

x + 4y ≥ 12

x ≥ 3, y ≥ 2

Solution:

Convert the given in equations into equations, we will get the following equations:

x + y = 8, x + 4y = 12, x = 3, y = 2

The area shows by x + y ≥ 8:

The line x + y = 8 connects the coordinate axes at A(8, 0) and B(0, 8) respectively. 

After connecting these points we will get the line x + y = 8.

Thus,

(0,0) does not assure the in equation x + y ≥ 8. 

Thus,

The region in xy plane which does not contain the origin represents the solution set of the in equation x + y ≥ 8.

The area shown by x + 4y ≥ 12:

The line x + 4y = 12 connects the coordinate axes at C(12, 0) and D(0, 3) respectively. 

After connecting these points we will get the line x + 4y = 12.

Thus,

(0,0) assure the in equation x + 4y ≥ 12. 

Thus,

the area in xy plane which having the origin represents the solution set of the in equation x + 4y ≥ 12.

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis.x ≥ 3 is the area to the right of the line

x = 3.

The line y = 2 is the line that passes through the point (0, 12) and is parallel to X axis.y ≥ 2 is the area above the line y = 2. 

The corner points of the suitable region are E(3, 5) and F(6, 2).
 

The values of Z at these corner points are as follows.

Corner pointZ = 2x + 4y
E(3, 5)2 × 3 + 4 × 5 = 26
F(6, 2)2 × 6 + 4 × 2 = 20

Therefore, 

The minimum value of Z is 20 at the point F(6, 2). 

Hence, x = 6 and y =2 is the best solution of the given LPP.

Therefore, 

The best value of Z is 20.

Question 11. Minimize Z = 5x + 3y

Subject to

2x + y ≥ 10

x + 3y ≥ 15

x ≤ 10

y ≤ 8

x, y ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

2x + y = 10, x + 3y = 15, x = 10, y = 8

Area shown by 2x + y ≥ 10:

The line 2x + y = 10 meets the coordinate axes at A(5, 0) and B(0, 10) respectively. 

After connecting these points we will get the line 2x + y = 10.

Thus,

(0,0) does not assure the in equation 2x + y ≥ 10. 

Thus,

The area in xy plane which does not have the origin represents the solution set of the in equation 2x + y ≥ 10.

The area represented by x + 3y ≥ 15:

The line x + 3y = 15 connects the coordinate axes at C(15, 0) and D(0, 5) respectively. 

After connecting these points we will get the line x + 3y = 15.

Thus,

(0,0) assure the in equation x + 3y ≥ 15.  

The area in xy plane which does not have the origin represents the solution set of the in equation x + 3y ≥ 15.

The line x = 10 is the line that passes through the point (10, 0) and is parallel to Y axis.x ≤ 10 is the area to the left of the line 

x = 10.

The line y = 8 is the line that passes through the point (0, 8) and is parallel to X axis.y ≤ 8 is the area below the line y = 8.

The area shows by x ≥ 0 and y ≥ 0:

Thus, 

All point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shows by the in equations x ≥ 0 and y ≥ 0.

The suitable area determined by the system of constraints, 2x + y ≥ 10, x + 3y ≥ 15, x ≤ 10, y ≤ 8, x ≥ 0 and y ≥ 0 are as 

follows.

The corner points of the suitable region are 

E(3, 4),

H(10,\ \frac{5}{3})

F(10, 8) and 

G(1, 8).

The values of Z at these corner points are as follows.

\frac{5}{3}

Hence, 

The minimum value of Z is 27 at the point F(3, 4). 

Therefore, 

x = 3 and y =4 is the best solution of the given LPP.

Hence, the best value of Z is 27.

Question 12.  Minimize Z = 30x + 20y

Subject to

x + y ≤ 8

x + 4y ≥ 12

5x + 8y = 20

x, y ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

x + y = 8, x + 4y = 12, x = 0 and y = 0

5x + 8y = 20 is already an equation.

The area shows by x + y ≤ 8:

The line x + y = 8 connects the coordinate axes at A(8, 0) and B(0, 8) respectively. 

After connecting these points we will get the line x + y = 8.

Thus,

(0,0) assures the in equation x + y ≤ 8. 

Thus,

The area in xy plane which have the origin represents the solution set of the in equation x + y ≤ 8.

The area shows by x + 4y ≥ 12:

The line x + 4y = 12 connects the coordinate axes at C(12, 0) and D(0, 3) respectively. 

After connecting these points we will get the line x + 4y = 12.

Thus, 

(0,0) assure the in equation x + 4y ≥ 12. 

Thus,

The area in xy plane which does not have the origin shows the solution set of the in equation x + 4y ≥ 12.

The line 5x + 8y = 20 is the line that passes through E(4, 0) and F (0,\ \frac{5}{2})    .

The area shows by x ≥ 0 and y ≥ 0:

Thus, 

All point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shows by the in equations x ≥ 0 and y ≥ 0.

The suitable area determined by the system of constraints, x + y ≤ 8, x + 4y ≥ 12, 5x + 8y = 20, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the suitable area are B(0,8), D(0,3), G \left(\frac{20}{3},\ \frac{4}{3}\right)    .
The values of Z at these corner points are as follows.

\left(\frac{20}{3},\ \frac{4}{3}\right)

Hence, 

The minimum value of Z is 60 at the point D(0,3).  

Therefore, 

x = 0 and y =3 is the best solution of the given LPP.

Therefore, 

The best value of Z is 60.

Question 13.  Maximize Z = 4x + 3y

Subject to

3x + 4y ≤ 24

8x + 6y ≤ 48

           x ≤ 6000

           x ≤ y

       x, y ≥ 0

Solution:

Here, we need to maximize Z = 4x + 3y

Convert the given in equations into equations, we will get the following equations:

3x + 4y = 24, 8x + 6y = 48, x = 5 , y = 6, x = 0 and y = 0.

The line 3x + 4y = 24 connects the coordinate axis at A(8, 0) and B(0,6). 

Connects these points to get the line 3x + 4y = 24.

Thus, 

(0, 0) assure the in equation 3x + 4y ≤ 24.

Thus, 

The area in xy-plane that have the origin represents the solution set of the given equation.

The line 8x + 6y = 48 connects the coordinate axis at C(6, 0) and D(0,8). 

Connect these points to get the line 8x + 6y = 48.

Thus, 

(0, 0) assure the in equation 8x + 6y  ≤ 48. 

Thus, 

The area in xy-plane that have the origin represents the solution set of the given equation.

x = 5 is the line passing through x = 5 parallel to the Y axis.

y = 6 is the line passing through y = 6 parallel to the X axis.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the points in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations.

These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0), G(5, 0), F (5,\ \frac{4}{3}), E (\frac{24}{7},\ \frac{24}{7}) and B(0, 6).

The values of Z at these corner points are as follows.
 

(5,\ \frac{4}{3})

Here we can see that the maximum value of the objective function Z is 24 which is at F (5,\ \frac{4}{3}) and E (\frac{24}{7},\ \frac{24}{7}).

Therefore, 

The best value of Z is 24.

Question 14. Minimize Z = x – 5y + 20

Subject to

    x – y ≥ 0

-x + 2y ≥ 2

          x ≥ 3

          y ≤  4

      x, y ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

x − y = 0, − x + 2y = 2, x = 3, y = 4, x = 0 and y = 0. 

The area shown by x − y ≥ 0 or x ≥ y:

The line x − y = 0 or x = y passes through the origin.The area to the right of the line x = y will assure the given in equation.

Now we will check by taking an example like if we take a point (4, 3) to the right of the line x = y .

Here, x ≥ y.

Thus, 

It assure the given in equation.  

Take a point (4, 5) to the left of the line x = y. Here, x ≤ y. That means it does not assure the given in equation.
 
The area shown by − x + 2y ≥ 2:

The line − x + 2y = 2 connects the coordinate axes at A(−2, 0) and B(0, 1) respectively. 

After connecting these points we will get the line − x + 2y = 2.

Thus, 

(0,0) does not assure the in equation − x + 2y ≥ 2. 

Thus,

The area in xy plane which does not have the origin represents the solution set of the in equation − x + 2y ≥ 2 .

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. x ≥ 3 is the area to the right of the 

line x = 3.

The line y = 4 is the line that passes through the point (0, 4) and is parallel to X axis. y ≤ 4 is the area below the line y = 4.
 
The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All point in the first quadrant assure these in equations. 

Thus, the first quadrant is the area shown by the in equations x ≥ 0 and y ≥ 0.

The suitable area determined by the system of constraints x − y ≥ 0,− x + 2y ≥ 2, x ≥ 3, y ≤ 4, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the suitable area are C (3,\ \frac{5}{2}), D(3, 3), E(4, 4) and F(6, 4).

The values of Z at these corner points are as follows.

(3,\ \frac{5}{2})

Hence, 

The minimum value of Z is 4 at the point E(4, 4). 

Therefore, 

x = 4 and y = 4 is the best solution of the given LPP.

Hence, 

The best value of Z is 4.

Question 15.  Maximize Z = 3x + 5y

Subject to

x + 2y ≤ 20

  x + y ≤ 15

         y ≤ 15

     x, y ≥ 0

Solution:

Here we need to maximize Z = 3x + 5y

Convert the given in equations into equations, we will get the following equations:

x + 2y = 20, x + y = 15, y = 5, x = 0 and y = 0.

The line x + 2y = 20 connects the coordinate axis at A(20, 0) and B(0,10). 

Connect these points to get the line x + 2y = 20.

Thus, 

(0, 0) assume the in equation x + 2y ≤ 20. 

Thus, 

The area in xy-plane that have the origin represents the solution set of the given equation.

The line x + y = 15 connect the coordinate axis at C(15, 0) and D(0,15). 

Connect these points to get the line x + y = 15.

Thus, 

(0, 0) assume the in equation x + y ≤ 15. 

Thus, 

The area in xy-plane that have the origin represents the solution set of the given equation.

y = 5 is the line passing through (0, 5) and parallel to the X axis. The area below the line y = 5 will assure the given in equation.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations.

These lines are drawn using a suitable scale.

The corner points of the suitable area are O(0, 0), C(15, 0), E(10, 5) and F(0, 5)

The values of Z at these corner points are as follows.
 

Corner pointZ = 3+ 5y
O(0, 0)3 × 0 + 5 × 0 = 0
 C(15, 0)3 × 15 + 5 × 0 = 45
E(10, 5)3 × 10 + 5 × 5 = 55
F(0, 5)3 × 0 + 5 × 5 = 25

Here we can see that the maximum value of the objective function Z is 55 which is at E(10, 5).

Therefore, 

The best value of Z is 55.

Question 16. Minimize Z = 3x1 + 5x2

Subject to

x1 + x2 ≥ 3

x1 + x2 ≥ 2

x1, x2 ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

x1 + 3x2 = 3, x1 + x2 = 2, x1 = 0 and x2 = 0

The area shown by x1 + 3x2 ≥ 3 :

The line x1 + 3x2 = 3 connects the coordinate axes at A(3, 0) and B(0, 1) respectively. 

After connecting these points we will get the line x1 + 3x2 = 3.

Thus,

(0,0) does not assure the in equation x1 + 3x2 ≥ 3.

Thus,

The area in the plane which does not have the origin represents the solution set of the in equation x1 + 3x2 ≥ 3.

The area shown by x1 + x2 ≥ 2:

The line x1 + x2 = 2 connects the coordinate axes at C(2, 0) and D(0, 2) respectively. 

After connecting these points we will get the line x1 + x2 = 2.

Thus, 

(0,0) does not assure the in equation x1 + x2 ≥ 2. 

Thus,

The area having the origin represents the solution set of the in equation x1 + x2 ≥ 2.

The area shown by x1 ≥ 0 and x2 ≥ 0:

Hence, 

All the points in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x1 ≥ 0 and x2 ≥ 0.

The suitable area determined by the system of constraints, x1 + 3x2 ≥ 3 , x1 + x2 ≥ 2, x1 ≥ 0, and x2 ≥ 0, are as follows.

The corner points of the suitable area are O(0, 0), B(0, 1), E (\frac{3}{2},\ \frac{1}{2}) and C(2, 0).

The values of Z at these corner points are as follows.

(\frac{3}{2},\ \frac{1}{2})

Hence, 

The minimum value of Z is 0 at the point O(0, 0). 

Hence, 

x1 = 0 and x2 = 0 is the best solution of the given LPP.

Therefore, 

The best value of Z is 0.

Question 17.  Maximize Z = 2x + 3y

Subject to

      x + y ≥ 1

  10x + y ≥ 5

  x + 10y ≥ 1

         x, y ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

x + y = 1, 10x +y = 5, x + 10y = 1, x = 0 and y = 0

The area shown by x + y ≥ 1:

The line x + y = 1 connects the coordinate axes at A(1, 0) and B(0,1) respectively. 

After connecting these points we will get the line x + y = 1.

Thus,

(0,0) does not assure the in equation x + y ≥ 1. 

Thus,

The area in xy plane which does not have the origin represents the solution set of the in equation x + y ≥ 1.

The area shown by 10x +y ≥ 5:

The line 10x +y = 5 connect the coordinate axes at C (\frac{1}{2},\ 0)    and D(0, 5) respectively. 

After connecting these points we will get the line 10x +y = 5.

Thus,

(0,0) does not assure the in equation 10x +y ≥ 5. 

Thus, the area which does not have the origin represents the solution set of the in equation 10x +y ≥ 5.

The area shown by x + 10y ≥ 1:

The line x + 10y = 1 connect the coordinate axes at A(1, 0) and F (0,\ \frac{1}{10})    respectively. 

After connecting these points we will have the line x + 10y = 1.

Thus, 

(0,0) does not assure the in equation x + 10y ≥ 1. 

Thus, the area which does not have the origin represents the solution set of the in equation x + 10y ≥ 1.

The area shown by x ≥ 0 and y ≥ 0:

Here, 

All the points in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints x + y ≥ 1, 10x +y ≥ 5, x + 10y ≥ 1, x ≥ 0, and y ≥ 0, are as follows.

The suitable area is unbounded.

Hence, 

The maximum value is infinity i.e. the solution is unbounded.

Question 18. Maximize Z = -x1 + 2x2

Subject to

  -x1 + 3x2 ≤ 10

     x1 + x2 ≤ 6

       x1 – x2 ≤ 2

        x1, x2 ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

−x1 + 3x2 = 10, x1 + x2 = 6, x1 + x2 = 2, x1 = 0 and x2 = 0

The area shown by −x1 + 3x2 ≤ 10:

The line −x1 + 3x2 = 10 coincide the coordinate axes at A(−10, 0) and B (0,\ \frac{10}{3})    respectively. 

After connecting these points we will get the line −x1 + 3x2 = 10.

Thus, 

(0,0) satisfies the in equation −x1 + 3x2 ≤ 10 .

Thus,

The area region in the plane which have the origin shows the solution set of the in equation

−x1 + 3x2 ≤ 10.

The area shown by x1 + x2 ≤ 6:

The line x1 + x2 = 6 connects the coordinate axes at C(6, 0) and D(0, 6) respectively. 

After connecting these points we will get the line x1 + x2 = 6.

Thus, 

(0,0) assure the in equation x1 + x2 ≤ 6. 

Thus,

The area having the origin represents the solution set of the in equation x1 + x2 ≤ 6.

The area shown by x1− x2 ≤ 2:

The line x1 − x2 = 2 coincide the coordinate axes at E(2, 0) and F(0, −2) respectively. 

After connecting these points we will get the line x1 − x2 = 2.

Thus,

(0,0) assure the in equation x1− x2 ≤ 2. 

Thus,

The area having the origin represents the solution set of the in equation x1− x2 ≤ 2.

The area shown by x1 ≥ 0 and x2 ≥ 0:

Hence, 

All the points in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the region shown by the in equations x1 ≥ 0 and x2 ≥ 0.

The suitable area determined by the system of constraints, −x1 + 3x2 ≤ 10, x1 + x2 ≤ 6, x1− x2 ≤ 2, x1 ≥ 0, and x2 ≥ 0, are as follows.

The corner points of the assure area are O(0, 0), E(2, 0), H(4, 2), G(2, 4) and B (0,\ \frac{10}{3})   .

The values of Z at these corner points are as follows.

(0,\ \frac{10}{3})

Here we can see that the maximum value of the objective function Z is \frac{20}{3} which is at B (0,\ \frac{10}{3}).

Question 19. Maximize Z = -x + y

Subject to

-2x + y ≤ 1

         x  ≤ 2

    x + y ≤ 3

       x, x ≥ 0

Solution:

Here we need to maximize Z = x + y

Convert the given in equations into equations, we will get the following equations:

−2x + y = 1, x = 2, x + y = 3, x = 0 and y = 0.

The line −2x + y = 1 coincide the coordinate axis at A (\frac{-1}{2},\ 0)    and B(0, 1). 

Now connect these points to get the line −2x + y = 1 .

Thus, 

(0, 0) assure the in equation −2x + y ≤ 1. 

Thus, 

The area in xy-plane that have the origin represents the solution set of the given equation.

x = 2 is the line passing through (2, 0) and parallel to the Y axis.

The area below the line x = 2 will assure the given in equation.

The line x + y = 3 coincide the coordinate axis at C(3, 0) and D(0, 3). 

Connect these points to get the line x + y = 3.

Thus, 

(0, 0) assure the in equation x + y ≤ 3. 

Thus, 

The area in xy-plane that have the origin shows the solution set of the given equation.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the points in the first quadrant assures these in equations. 

Thus, 

The first quadrant is the area shown by the in equations.

These lines are drawn using a satisfactory scale.

The corner points of the suitable area are O(0, 0),G(2, 0), E(2, 1) and F (\frac{2}{3},\ \frac{7}{3})

The values of Z at these corner points are as follows.
 

(\frac{2}{3},\ \frac{7}{3})

Here we can see that the maximum value of the objective function Z is 3 which is at E(2, 1) and F (\frac{2}{3},\ \frac{7}{3})   .

Therefore, 

The best value of Z is 3.

Question 20.  Maximize Z = -3x1 + 4x2

Subject to

     x1 – x2 ≤ -1

   x1 + x2  ≤ 0

       x1, x2 ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

x1 − x2 = −1, −x1 + x2 = 0, x1 = 0 and x2 = 0

The area shown by x1 − x2 ≤ −1:

The line x1 − x2 = −1 coincide the coordinate axes at A(−1, 0) and B(0, 1) respectively. 

After connecting these points we will get the line x1 − x2 = −1.

Thus, 

(0,0) does not assures the in equation x1 − x2 ≤ −1 .

Thus,

The area in the plane which does not have the origin shows the solution set of the in equation x1 − x2 ≤ −1.

The area shown by −x1 + x2 ≤ 0 or x1 ≥ x2:

The line −x1 + x2 = 0 or x1 = x2 is the line passing through (0, 0).

The area to the right of the line x1 = x2 will assure the given in equation −x1 + x2 ≤ 0.

If we mark a point (1, 3) to the left of the line x1 = x2

Here, 1≤3 which is not assuring the in equation x1 ≥ x2

Hence, 

The area to the right of the line x1 = x2 will assure the given in equation −x1 + x2 ≤ 0.

The area shown by x1 ≥ 0 and x2 ≥ 0:

Thus, 

All the points in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x1 ≥ 0 and x2 ≥ 0.

The suitable area determined by the system of constraints, x1 − x2 ≤ −1, −x1 + x2 ≤ 0, x1 ≥ 0, and x2 ≥ 0, are as follows.

Here we can see that the suitable area of the given LPP does not exist.

Question 21: Maximize Z = 3x + 3y, if possible,

Subject to the constraints

x − y ≤ 1

x + y ≥ 3 

x, y ≥ 0

Solution:

Convert the given in equations into equations, we will get the following equations:

x − y = 1, x + y = 3, x = 0 and y = 0

The area shown by x − y ≤ 1:

The line x − y = 1 coincide the coordinate axes at A(1, 0) and B(0, −1) respectively. 

After connecting these points we will get the line x − y = 1.

Thus,

(0,0) assure the in equation x + y ≤ 8. 

Thus,

The area in xy plane which have the origin shown the solution set of the in equation x − y ≤ 1.

The area shown by x + y ≥ 3:

The line x + y = 3 coincide the coordinate axes at C(3, 0) and D(0, 3) respectively. 

After connecting these points we will get the line x + y = 3.

Thus,

(0,0) assure the in equation x + y ≥ 3. 

Thus,

The area in xy plane which does not have the origin show the solution set of the in equation x + y ≥ 3.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the points in the first quadrant assure these in equations. Thus, the first quadrant is the area shown by the in equations 

x ≥ 0 and y ≥ 0.

The suitable area determined by the system of constraints x − y ≤ 1, x + y ≥ 3, x ≥ 0 and y ≥ 0 are as follows.

The suitable area is unbounded.

We will get the maximum value at infinity.

Hence,

Maximum value will be infinity i.e. the solution is unbounded.

Question 22. Show the solution zone of the following inequalities on a graph paper:

5x + y ≥ 10   

  x + y ≥ 6

x + 4y ≥ 12

         x ≥ 0, 

         y ≥ 0

Find x and y for which 3x + 2y is minimum subject to these inequalities. Use a graphical method.

Solution:

Convert the given in equations into equations, we will get the following equations:

5x + y = 10, x +y = 6, x + 4y = 12, x = 0 and y = 0

The area shown by 5x + y ≥ 10:

The line 5x + y = 10  coincide the coordinate axes at A(2, 0) and B(0, 10) respectively. 

After connecting these points we will get the line 5x + y = 10.

Thus, 

(0,0) does not assure the in equation 5x + y ≥ 10. 

Thus,

The area in xy plane which does not have the origin shows the solution set of the in equation 5x + y ≥ 10.

The area shown by x +y  ≥ 6:

The line x +y = 6 coincide the coordinate axes at C(6,0) and D(0, 6) respectively. 

After connecting these points we will get the line 2x +3y = 30.

Thus, 

(0,0) does not assure the in equation x +y  ≥ 6. 

Thus,

The area which does not have the origin shows the solution set of the in equation 2x +3y ≥ 30.

The area shown by x + 4y ≥ 12

The line x + 4y = 12 coincide the coordinate axes at E(12, 0) and F(0, 3) respectively. 

After connecting these points we will get the line x + 4y = 12.

Thus, 

(0,0) does not assure the in equation x + 4y ≥ 12. 

Thus,

The area which does not have the origin shows the solution set of the in equation x + 4y ≥  12.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the points in the first quadrant assure these in equations. Thus, the first quadrant is the area shown by the inequations 

x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints 5x + y ≥ 10, x +y  ≥ 6,x + 4y ≥ 12, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the suitable area are B(0, 10), G(1,5), H(4,2) and E(12,0).

The values of Z at these corner points are as follows.

Corner pointZ = 3+ 2y
B(0, 10)3 × 0 + 3 × 10 = 30
G(1,5)3 × 1 + 2 × 5 = 13
H(4,2) 3 × 4 + 2 × 2 = 16
E(12,0)3 × 12 + 2 × 0 = 36

Hence, 

The minimum value of Z is 13 at the point G(1,5). 

Therefore, 

x = 1 and y = 5 is the best solution of the given LPP.

Therefore, the best value of Z is 13.

Question 23: Find the maximum and minimum value of 2x + y subject to the constraints:

x + 3y ≥ 6, 

x − 3y ≤ 3, 

3x + 4y ≤ 24, 

− 3x + 2y ≤ 6, 

5x + y ≥ 5,

 x, y ≥ 0.

Solution:

Convert the given in equations into equations, we will get the following equations:

x + y = 4, x + y = 3, x − 2y = 2, x = 0 and y = 0.

The line x + 3y = 6 coincide the coordinate axis at A(6, 0)A6, 0 and B(0, 2). 

Connect these points to get the line x + 3y = 6.

Thus, 

(0, 0) does not assure the in equation x + 3y ≥ 6. Thus, the area in xy-plane that does not have the origin represents the

solution set of the given equation.

The line x − 3y = 3 coincide the coordinate axis at C(3, 0) and D(0, −1). 

Connect these points to get the line x − 3y = 3.

Thus, 

(0, 0) assure the in equation x − 3y ≤ 3. Thus, the area in xy-plane that have the origin represents the solution set of the given

equation.

The line 3x + 4y = 24 meets the coordinate axis at E(8, 0) and F(0, 6). Join these points to get the line 3x + 4y = 24.

Thus, 

(0, 0) assure the in equation 3x + 4y ≤ 24. Thus, the area in xy-plane that have the origin shows the solution set of the given 

equation.

The line −3x + 2y = 6 coincide the coordinate axis at G(−2, 0) and H(0, 3). 

Connect these points to get the line −3x + 2y = 6.

Thus, (0, 0) assure the in equation −3x + 2y ≤ 6. 

Thus, 

The area in xy-plane that have the origin represents the solution set of the given equation.

The line 5x + y = 5 coincide the coordinate axis at I(1, 0)I1, 0 and J(0, 5). 

Connect these points to get the line 5x + y = 5.

Thus, 

(0, 0) does not assure the in equation 5x + y ≥ 5. Thus, the area in xy-plane that does not have the origin have the 

solution set of the given equation.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the points in the first quadrant assure these inequations. Thus, the first quadrant is the area shown by the in equations.

These lines are drawn using a satisfactory scale.

The corner points of the suitable area

 are P (\frac{4}{13},\ \frac{45}{13})      , K (\frac{4}{3},\ 5)     , L (\frac{84}{13},\ \frac{15}{13})     , M (\frac{9}{2},\ \frac{1}{2})     ,

(\frac{9}{14},\ \frac{25}{14})      

The values of Z at these corner points are as follows.
 

(\frac{4}{13},\ \frac{45}{13})

Here, we can see that the minimum value of the objective function Z is \frac{43}{14}      which is at N (\frac{9}{14},\ \frac{25}{14})       

and maximum value of the objective function is \frac{183}{13}      which is at L (\frac{84}{13},\ \frac{15}{13})     .

Question 24: Find the minimum value of 3x + 5y subject to the constraints

− 2x + y ≤ 4, 

x + y ≥ 3, 

x − 2y ≤ 2, 

x, y ≥ 0.

Solution:

Convert the given in equations into equations, we will get the following equations:

−2x + y = 4, x + y = 3, x − 2y = 2, x = 0 and y = 0.

The line −2x + y = 4 coincide the coordinate axis at A(−2, 0)A-2, 0 and B(0, 4). 

Connect these points to obtain the line −2x + y = 4.

Thus, (0, 0) assure the in equation −2x + y ≤ 4. So, the area in xy-plane that have the origin represents the solution set of the 

given equation.

The line x + y = 3 coincide the coordinate axis at C(3, 0) and D(0, 3). 

Connect these points to get the line x + y = 3.

Thus, 

(0, 0) does not assure the in equation x + y ≥ 3. Thus, the area in xy-plane that does not have the origin shows the solution set 

of the given equation.

The line x − 2y = 2 coincide the coordinate axis at E(2, 0) and F(0, −1). 

Connect these points to get the line x − 2y = 2.

Thus, 

(0, 0) assure the in equation x − 2y ≤ 2. Thus, the area in xy-plane that have the origin represents the solution set of the given 

equation.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the points in the first quadrant assure these in equations. Thus, the first quadrant is the area shown by the in equations.

These lines are drawn using a satisfactory scale.

The corner points of the suitable area are B(0, 4), D(0, 3) and G (\frac{8}{3},\ \frac{1}{3})     .

The values of Z at these corner points are as follows.
 

(\frac{8}{3},\ \frac{1}{3})

Here we can see that the minimum value of the objective function Z is \frac{29}{3}      which is at G (\frac{8}{3},\ \frac{1}{3})     .

Therefore, 

The best value of Z is \frac{29}{3}     .

Question 25: Solved the following linear programming problem graphically:

Maximize Z = 60x + 15y

Subject to constraints

  x + y ≤ 50

3x + y ≤ 90

      x, y≥0 

Solution:

Here we have to maximize Z = 60x + 15y

Convert the given in equations into equations, we get the following equations:

x + y = 50, 3x + y = 90, x = 0 and y = 0

The area shown by x + y  ≤ 50:

The line x + y  = 50 coincide the coordinate axes at A(50,0) and B(0,50) respectively. 

After connecting these points we will get line 3x + 5y = 15.

Thus,

(0,0) assure the in equation x + y  ≤ 50. Thus, the area having the origin shows the solution set of the in equation x + y  ≤ 50.

The area shown by 3x + y ≤ 90:

The line 3x + y = 90 coincide the coordinate axes at C(30, 0) and D(0, 90) respectively. 

After connecting these points we will get the line 3x + y = 90.

Thus,

(0,0) assure the in equation 3x + y ≤ 90. Thus, the area having the origin shows the solution set of the in equation 3x + y ≤ 90.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the points in the first quadrant assure these in equations. Thus, the first quadrant is the area shown by the in equations 

x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints, x + y  ≤ 50, 3x + y ≤ 90, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the suitable area are O(0, 0), C(30, 0), E(20, 30) and B(0, 50).

The values of Z at these corner points are as follows.

Corner pointZ = 60+ 15y
O(0, 0)60 × 0 + 15 × 0 = 0
C(30, 0)60 × 30 + 15 × 0 = 1800
E(20, 30)E20, 3060 × 20 + 15 × 30 =1650
B(0, 50)60 × 0 + 15 × 50 = 750

Hence, 

The maximum value of Z is 1800 at the point (30, 0) at the point 30, 0.

Therefore,

x = 30 and y = 0 is the best solution of the given LPP.

Therefore, 

The best value of Z is 1800.

Question 26. Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:

2x + 4y ≤ 8

3x + y ≤ 6

x + y ≤ 4

x ≥ 0, y ≥ 0                      [CBSE 2015]

Solution:

Here, 

Maximize Z = 2x + 5y subject to the constraints

2x + 4y ≤ 8

3x + y ≤ 6

x + y ≤ 4

x ≥ 0, y ≥ 0

After converting the in equations into equations, we will get the following equations of straight lines:

2x + 4y = 8, 3x + y = 6, x + y = 4

The line 2x + 4y = 8 coincide the coordinate axes at (4, 0) and (0, 2).

The line 3x + y = 6 coincide the coordinate axes at (2, 0) and (0, 6).

The line x + y = 4 coincide the coordinate axes at (4, 0) and (0, 4).

The suitable area determined by the given constraints can be diagrammatically shown as,

The coordinates of the corner points of the suitable area are O(0, 0), A(0, 2), B (\frac{8}{5},\ \frac{6}{5})      and C(2, 0).

The value of the objective function at these points are given in the following table.
 

(\frac{8}{5},\ \frac{6}{5})

Therefore, 

The maximum value of Z is 10 at x = 0, y = 2.

Question 27: Solve the following LPP graphically:

Maximize Z = 20x + x + 10y 

Subject to the following constraints

x + 2y ≤ 28

3x + y ≤ 24 

x ≥ 2  

x,y ≥  0

Solution:

Here given constraints are

x + 2y ≤ 28     
   
3x + y ≤ 24  
       
x ≥ 2       
          
x, y ≥ 0              

Converting the given in equations into equations, 

we will get

x + 2y = 28, 3x + y = 24, x = 2, x = 0 and y = 0

These lines are drawn on the graph and the shaded region ABCD shown the suitable area of the given LPP.

It can be seen that the suitable area is bounded. 

The coordinates of the corner points of the suitable area are A(2, 13), B(2, 0), C(4, 12) and D(8, 0).

The values of the objective function, Z at these corner points are given in the following table:
 

Corner PointValue of the Objective Function Z = 20+ 10y
A(2, 13)Z = 20 × 2 + 10 × 13 = 170
B(2, 0)Z = 20 × 2 + 10 × 0 = 40
C(8, 0)Z = 20 × 8 + 10 × 0 = 160
D(4, 12)Z = 20 × 4 + 10 × 12 = 200 

As can be seen in the table, is maximum at =4 and y = 12 and the maximum value of is 200.

Therefore, 

The maximum value of Z is 200.

Question 28: Solve the following linear programming problem graphically:

Minimize z= 6x + 3y

Subject to the constraints:

4x + y ≥ 80

x + 5y ≥ 115

3x + 2y ≤ 150

x ≥ 0, y ≥ 0

Solution:

Here the given constraints are

4x + y ≥ 80     
   
x + 5y ≥ 115  
       
3x + 2y ≤ 150       
          
x, y ≥ 0              

Converting the given in equations into equations, 

We will get

4x + y = 80, x + 5y = 115, 3x + 2y = 150, x = 0 and y = 0

These lines are drawn on the graph and the shaded area ABC represents the suitable area of the given LPP.

It can be seen that the suitable area is bounded. 

The coordinates of the corner points of the suitable are A(2, 72), B(15, 20) and C(40, 15).

The values of the objective function, Z at these corner points are given in the following table:
 

Corner PointValue of the Objective Function Z = 6+ 3y
A(2, 72)Z = 6 × 2 + 3 × 72 = 228
B(15, 20)Z = 6 × 15 + 3 × 20 = 150
C(40, 15)Z = 6 × 40 + 3 × 15 = 285

As seen in the table, is minimum at =15 and y = 20 and the minimum value of is 150.

Therefore, 

The minimum value of Z is 150.

Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.3

  • Last Updated : 20 May, 2021

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Question 1. A diet of two foods F1 and F2 contains nutrients thiamine, phosphorus, and iron. The amount of each nutrient in each of the food(in milligrams per 25 grams) is given in the following table.

Nutrients/FoodF1F2
Thiamine0.250.10
Phosphorus0.751.50
Iron1.600.80

The minimum requirement of the nutrients in the diet is 1.00 mg of thiamine,7.50 mg of phosphorus, and 10.00 mg of iron. The cost of F1 is 20 paise per 25 grams while the cost of F2 is 15 paise per 25 grams. Find the minimum cost of the diet.

Solution:

Let 25x grams of food F1 and 25y grams of food F2 be used to fulfill the minimum requirement of thiamine, phosphorus and iron.

Given        

Nutrients/FoodF1F2
Thiamine0.250.10
Phosphorus0.751.50
Iron1.600.80

The minimum requirement of the nutrients in the diet are 1.00 mg of thiamine,7.50 mg of phosphorus and 10.00 mg of iron. Therefore,

0.25x+0.10y ≥ 1

0.75x+1.50y ≥7.5

1.6x+0.8y ≥ 10

Since the quantity cannot be negative 

⇒ x,y ≥ 0

The cost of F1 is 20 paise per 25 grams and the cost of F2 is 15 paise per 25 grams. 

Therefore the cost of 25x grams of food F1 and 25y grams of food F2 is (0.20x+0.15y) rupees.

Let us minimize z=0.20x+0.15y

subject to

0.25x+0.10y ≥ 1

0.75x+1.50y ≥7.5

1.6x+0.8y ≥ 10

x, y ≥ 0

First we will convert the given inequations to equations. we get,

0.25x+0.10y = 1

0.75x+1.50y =7.5

1.6x+0.8y = 10

x=0 and y=0

Region represented by 0.25x+0.10y ≥ 1:

The line 0.25x+0.10y =1 meets the coordinate axis at A(4, 0) and B(0, 10).

Clearly (0, 0) do not satisfy the inequation 0.25x+0.10y≥1.So the region in XY plane that does not contain the origin represents the solution set of given inequation.

Region represented by 0.75x+1.50y ≥7.5:

The line 0.75x+1.50y =7.5 meets the coordinate axis at C(10, 0) and D(0, 5).

Clearly (0, 0) do not satisfy the inequation 0.75x+1.50y≥7.5.So the region in XY plane that doe snot contain the origin represents the solution set of this inequation.

Region represented by 1.6x+0.8y ≥ 10:

The line 1.6x+0.8y = 10 meets the coordinate axis at E(25/4, 0) and F(0, 25/2).

Clearly (0, 0) do not satisfy the inequation 1.6x+0.8y ≥ 10.So the region in XY plane that does not contain the origin represents the solution set of given inequation.

Region represented by x,y ≥ 0:

In the region represented by x≥0 and y≥0, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are F(0, 12.5), G(5, 2.5), C(10, 0)

The value of objective function at these points are given in the following table 

PointsValue of z
F0.20(0)+0.15(12.5)=1.875
G0.20(5)+0.15(2.5)=1.375
C0.20(10)+00.15(0)=200

Thus, the minimum cost is at G which is 1.375 rupees.

Question 2. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals, and 1400 units of calories. Two foods A and B are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of A contains 200 units of vitamin,1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?

Solution:

Let the sick person takes x units and y units of food A and B respectively.

1 unit of food A costs Rs 4 and that of food B costs Rs 3.

Therefore, x units of food A costs Rs 4x and y units of food B costs Rs 3y.

Total cost =Rs (4x+3y) 

Let z=4x+3y

If one unit of A contains 200 units of vitamin and one unit of food B contains 100 units of vitamin.

Then, x units of food A and y units of food B contains (200x+100y) units of vitamin.

But a diet for a sick person must contain at least 4000 units of vitamin.

Thus 200x+100y≥4000

If one unit of A contains 1 unit of mineral and one unit of food B contains 2 units of mineral.

Then, x units of food A and y units of food B contains x+2y units of mineral.

But a diet for a sick person must contain at least 50 units of vitamin.

Thus, x+2y≥50

If one unit of A contains 40 calories and one unit of food B contains 40 calories.

Then x units of food A and y units of food B contains 40x+40y units of calories.

But a diet for a sick person must contain at least 1400 units of calories.

Thus, 40x+40y≥1400

Finally the quantities of food A and food B are nonnegative values.

So, x, y ≥ 0

Hence the required LPP is as follows:

Min z=4x+3y

subject to

200x+100y≥4000

x+2y≥50

40x+40y≥1400

 x, y ≥ 0

First we will convert the given inequations to equations. we get,

200x+100y=4000

x+2y=50

40x+40y=1400

x=0 and y=0

Region represented by 200x+100y≥4000:

The line 200x+100y=4000 meets the coordinate axis at A1(20, 0) and B1(0, 40) respectively.

Clearly (0, 0) does not satisfy inequation 200x+100y≥4000.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x+2y≥50:

The line x+2y=50 meets the coordinate axis at C1(50, 0) and D1(0, 25) respectively.

Clearly (0, 0) does not satisfy inequation x+2y≥50.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by 40x+40y≥1400:

The line 40x+40y=1400 meets the coordinate axis at E1(35, 0) and F1(0, 35) respectively.

Clearly (0, 0) does not satisfy inequation 40x+40y≥1400.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x≥ 0 and y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are B1(0, 40), G1(5, 30), C1(50, 0), H1(20, 15).

The value of objective function at these points are given in the following table

PointsValue of z
B14(0)+3(40)=120
G14(5)+3(30)=110
H14(20)+3(15)=125
C14(50)+3(0)=200

Thus the minimum cost is at G1 which is 110 rupees.

Question 3. To maintain one’s health, a person must fulfill certain minimum daily requirements for the following three nutrients: Calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:

 Food IFood IIMinimum daily requirement
Calcium10420
Protein5620
Calories2612
PriceRs 0.60per unitRs 1.00per unit 

Find the combination of food items so that the cost may be minimum.

Solution:

Let the person takes x units and y units of food I and II respectively.

Since, per unit of food I costs Rs 0.60 and that of food II costs Rs.1.00.

Therefore, x units of food I costs Rs 0.60x and y units of food II costs Rs1.00y.

Total cost per day=0.60x+1.00y

Let z=0.60x+1.00y

Since, each unit of food I contains 10 units of calcium.

Therefore, x units of food I contains 10x units of calcium.

Each unit of food II contains 4 units of Calcium.

So, y units of food II contains 4y units of calcium.

Thus, x units of food I and y units of food II contains (10x+4y) units of calcium.

But the minimum requirement is 20 units of calcium.

Thus, 10x+4y≥20

Since, each unit of food I contains 5 units of protein.

Therefore, x units of food I contains 5x units of protein.

Each unit of food II contains 6 units of protein.

So, y units of food II contains 6y units of protein.

Thus, x units of food I and y units of food II contains (5x+6y) units of protein.

But the minimum requirement is 20 units of protein.

Thus, 5x+6y≥20

Since, each unit of food I contains 2 units of Calories.

Therefore, x units of food I contains 2x units of Calories.

Each unit of food II contains 6 units of Calories.

So, y units of food II contains 6y units of Calories.

Thus, x units of food I and y units of food II contains (2x+6y) units of Calories.

But the minimum requirement is 12 units of Calories.

Thus, 2x+6y≥12

Finally the quantities of food I and food II are nonnegative values.

So, x, y ≥ 0

Hence the required LPP is as follows:

Min z=0.60x+1.00y

subject to

10x+4y≥20

5x+6y≥20

2x+6y≥12

x, y ≥ 0

First we will convert the given inequations to equations. we get,

10x+4y=20

5x+6y=20

2x+6y=12

x=0 and y=0

Region represented by 10x+4y≥20:

The line 10x+4y=20 meets the coordinate axes at A(2, 0) and B(0, 5) respectively.

Clearly (0, 0) does not satisfy inequation 10x+4y≥20.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by 5x+6y≥20:

The line 5x+6y=20 meets the coordinate axes at C(4, 0) and D(0, 10/3) respectively.

Clearly (0, 0) does not satisfy inequation 5x+6y≥20.

So the region in XY plane that does not contain the origin represents the solution set of this inequation

Region represented by 2x+6y≥12:

The line 2x+6y=12 meets the coordinate axes at E(6, 0) and F(0, 2) respectively.

Clearly (0, 0) does not satisfy inequation 2x+6y≥12.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x, y ≥ 0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are B(0, 5), G(1, 5/2), E(6, 0), H(8/3, 10/9).

The value of objective function at these points are given in the following table

PointsValue of z
B0.6(0)+5=5
G0.6(1)+5/2=3.1
H0.6(8/3)+(10/9)=2.711
E0.6(6)+0=3.6

From the table, we can see that minimum cost is at H(8/3, 10/9) and that value is Rs 2.71

Question 4. A hospital dietician wishes to find the cheapest combination of two foods, A and B that contains at least 0.5 milligrams of thiamine and at least 600 calories. Each unit of A contains 0.12 milligrams of thiamine and 100 calories, while each unit of B contains 0.10 milligrams of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?

Solution:

Let the dietician wishes to mix x units of food A and y units of food B

Therefore x,y≥0

The given information can be tabulated as follows

 Thiamine(mg)Calories(mg)
Food A0.12100
Food B0.1150
Minimumrequirement0.5600

According to question,

The constraints are

0.2x+0.1y≥0.5

100x+150y≥600

It is given that each food costs 10 paise per unit

Let the total cost be z

 z=10x+10y

Now the mathematical formulation of given linear programming problem is

0.2x+0.1y≥0.5

100x+150y≥600

Region represented by 0.12x+0.1y≥0.5:

The line 0.12x+0.1y=0.5 meets the coordinate axes at A1(25/6, 0) and B1(0, 5) respectively.

Clearly (0, 0) does not satisfy inequation 0.12x+0.1y≥0.5.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by 100x+150y≥600:

The line 100x+150y=600 meets the coordinate axes at C1(6, 0) and D1(0, 4) respectively.

Clearly (0, 0) does not satisfy inequation 100x+150y≥600.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are B1(0, 5), E1(15/8, 11/4), C1(6, 0).

The value of objective function at these points are given in the following table

Corner pointz=10x+10y
B150
E146.2
C160

The minimum of z is at E1(15/8, 11/4) and the value is Rs 46.2.

Hence the cheapest combination of foods will be 1.875 units of food A and 2.75 units of food B.

Question 5. A dietician mixes together 2 kinds of food in such a way that the mixture contains at least 6 units of vitamin A,7 units of vitamin B,11 units of vitamin C, and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:

 VitaminAVitaminBVitaminCVitaminD
Food X1112
Food Y2131

One kg of food X costs Rs 5 and one kg of Food Y costs Rs 8. Find the least cost of the mixture which will produce the desired diet.

Solution:

Let the dietician wishes to mix x kg of food X and y kg of food Y.

Therefore, x,y≥0

Given

 VitaminAVitaminBVitaminCVitaminD
Food X1112
Food Y2131

It is given that the mixture should contain at least 6 units of vitamin A,7 units of vitamin B ,11 units of vitamin C and 9 units of vitamin D.

Therefore, the constraints are

x+2y≥6

x+y≥7

x+3y≥11

2x+y≥9

It is given that cost of food X is Rs 5 per kg and cost of food Y is Rs 8 per kg.

Let the total cost be z.

Thus, z=5x+8y

Now the mathematical formulation of given linear programming problem is

Minimize Z=5x+8y

subject to

x+2y≥6

x+y≥7

x+3y≥11

2x+y≥9

First we will convert the given inequations to equations. we get

x+2y=6

x + y=7

x+3y=11

2x+y=9

x=0 and y=0

Region represented by x+2y≥6:

The line x+2y=6 meets the coordinate axes at A1(6, 0) and B1(0, 3) respectively.

Clearly (0, 0) does not satisfy inequation x+2y≥6.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x+y≥7:

The line x + y=7 meets the coordinate axes at C1(7, 0) and D1(0, 7) respectively.

Clearly (0, 0) does not satisfy inequation x+y≥7.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x+3y≥11:

The line x+3y=11 meets the coordinate axes at E1(11, 0) and F1(0, 11/3) respectively.

Clearly (0, 0) does not satisfy inequation x+3y≥11.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by 2x+y≥9:

The line 2x+y=9 meets the coordinate axes at G1(9/2, 0) and H1(0, 9) respectively.

Clearly (0, 0) does not satisfy inequation 2x+y≥9.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are H1(0, 9), I1(2, 5), E1(11, 0), J1(5, 2).

The value of objective function at these points are given in the following table.

Corner Pointz=5x+8y
H172
I150
J141
E155

The minimum value of z is at J1(5, 2) which is Rs 41.

Hence cheapest combination of foods will be 5 units of food X and 2 units of food Y.

Question 6. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost of diet that consists of mixture of these foods and also meets the mineral nutritional requirements.

Solution:

Let the dietician wishes to mix x units of food F1 and y units of food F2.

Clearly x,y≥0

The given information can be tabulated as follows

 VitaminAVitaminB
Food F134
Food F263
Minimum requirement80100

The constraints are

3x+6y≥80

4x+3y≥100

It is given that cost of food F1 is Rs 4 per unit and cost of food F2 is Rs 6 per unit.

Therefore cost of x units of food F1 and y units of food F2 is Rs 4x and Rs 6y respectively.

Let the total cost be z.

z=4x+6y

Now the mathematical formulation of given linear programming problem is

Minimize Z=4x+6y

subject to

3x+6y≥80

4x+3y≥100

 x,y≥0

First we will convert the given inequations to equations. we get

3x+6y=80

4x+3y=100

x=0 and y=0

Region represented by 3x+6y≥80:

The line 3x+6y=80 meets the coordinate axes at A(80/3, 0) and B(0, 40/3) respectively.

Clearly (0, 0) does not satisfy inequation 3x+6y≥80.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by 4x+3y≥100:

The line 4x+3y=100 meets the coordinate axes at C(25, 0) and D(0, 100/3) respectively.

Clearly (0, 0) does not satisfy inequation 4x+3y≥100.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are D(0, 100/3), E(24, 4/3), A(80/3, 0).

The value of objective function at these points are given in the following table.

Corner Pointz=4x+6y
D(0, 100/3)200
E(24, 4/3)104
A(80/3, 0)320/3

The minimum value of z is at E(24, 4/3) and the value is Rs 104.

Question 7. Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kg and that rice contains 100 grams of protein and 30 milligrams of iron per kg, Find the minimum cost of producing this new cereal if bran costs Rs 5 per kg and rice costs Rs 4 per kg.

Solution:

Let the cereal contain x kg of bran and y kg of rice.

Therefore x,y≥0

The given information can be tabulated as follows

 Protein(gms)Iron(milligrams)
Bran8040
Rice10030
Minimumrequirement8836

Bran and Rice contains at least 88 grams of protein and at least 36 milligrams of iron.

Thus the constraints are

80x+100y≥88

40x+30y≥36

It is given that bran costs Rs 5 per kg and rice costs Rs 4 per kg. Therefore cost of x kg of bran and y kg of rice is Rs 5x and Rs 4y respectively.

Hence, total profit is Rs(5x+4y).

Let the total cost be z

Thus, z=5x+4y

Now the mathematical formulation of given linear programming problem is

Minimize Z=5x+4y

subject to

80x+100y≥88

40x+30y≥36

x,y≥0

First we will convert the given inequations to equations. we get

80x+100y=88

40x+30y=36

x=0 and y=0

Region represented by 80x+100y≥88:

The line 80x+100y=88 meets the coordinate axes at A(1.1, 0) and B(0, 0.88) respectively.

Clearly (0, 0) does not satisfy inequation 80x+100y≥88.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by 40x+30y≥36:

The line 40x+30y=36 meets the coordinate axes at C(0.9, 0) and D(0, 1.2) respectively.

Clearly (0, 0) does not satisfy inequation 40x+30y≥36.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are D(0, 1.2), E(0.6, 0.4), A(1.1, 0).

The value of objective function at these points are given in the following table.

Corner Pointz=5x+4y
D(0, 1.2)4.8
E(0.6, 0.4)4.6
A(1.1, 0)5.5

The minimum value of z is Rs 4.6 at E(0.6, 0.4)

Question 8. A Wholesale dealer deals in two kinds A and B of the mixture of nuts. Each kg of mixture A contains 60 grams of almonds,30 grams of cashew nuts, and 30 grams of hazelnuts. Each kg of mixture B contains 30 grams of almonds,60 grams of cashew nuts, and 180 grams of hazelnuts. The remainder of both mixtures is per nuts. The dealer is contemplating using mixtures A and B to make a bag that will contain at least 240 grams of almonds,300 grams of cashew nuts, and 540 grams of hazelnuts. Mixture A costs Rs 8 per kg and mixture B costs Rs 12 per kg. Assuming that mixtures A and B are uniform, use the graphical method to determine the number of kg of each mixture which he should use to minimize the cost of the bag.

Solution:

Let x kg of kind A and y kg of kind B were used.

Since quantity cannot be negative.

Therefore x,y≥0

The given information can be tabulated as follows

NutAlmonds(grams)Cashew nuts(grams)
A(x)6030
B(x)3060
Availability240300

Thus the constraints are

60x+30y≥240

30x+60y≥300

30x+180y≥540

Mixture A costs Rs 8 per kg and mixture B costs Rs 12 per kg

Total cost= 8x+12y

Let the total cost be z

Thus, z=8x+12y

Now the mathematical formulation of given linear programming problem is

Minimize Z=8x+12y

subject to

2x+y≥8

x+2y≥10

x+6y≥18

x,y≥0

First we will convert the given inequations to equations. we get

2x+y=8

x+2y=10

x+6y=18

x=0 and y=0

Region represented by 2x+y≥8:

The line 2x+y=8 meets the coordinate axes at A1(4, 0) and B1(0, 8) respectively.

Clearly (0, 0) does not satisfy inequation 2x+y≥8.

So the region in XY plane that does not contain the origin represents the solution set of this equation

Region represented by x+2y≥10:

The line x+2y=10 meets the coordinate axes at C1(10, 0) and D1(0, 5) respectively.

Clearly (0, 0) does not satisfy inequation x+2y≥10.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x+6y≥18:

The line x+6y=18 meets the coordinate axes at E1(18, 0) and F1(0, 3) respectively.

Clearly (0, 0) does not satisfy inequation x+6y≥18.

So the region in XY plane that does not contains the origin represents the solution set of this equation

Region represented by x,y≥0:

Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are B1(0, 8), G1(2, 4), H1(6, 2) and E1(18, 0).

The value of objective function at these points are given in the following table.

Corner Pointz=8x+12y
B196
G164
H172
E1144

The minimum value of z is at G1(2, 4) and the value is Rs 64.

Thus the minimum cost is Rs 64 obtained when 2 units of kind A and 4 units of kind B nuts were used.

Question 9. One kind of cake requires 300 grams of flour and 15 grams of fat, another kind of cake requires 150 gms of flour and 30 gms of fat. Find the maximum number of cakes that can be made from 7.5 kg of flour and 600 gms of fat, assuming that there is no shortage of other ingredients used in making the cake. Make it as an LPP and solve it graphically.

Solution:

Let there be x cakes of first kind and y cakes of second kind.

Therefore x,y≥0

The given information can be tabulated as follows

 Flour(gm)Fat(gm)
Cakes of first kind30015
Cakes of second kind15030
Availability7500600

Thus the constraints are

300x+150y≤7500

15x+30y≤600

Total number of cakes that can be made =x + y

Let the total number of cakes be z.

Thus z=x + y

Now the mathematical formulation of given linear programming problem is

Maximize z=x + y

subject to

300x+150y≤7500

15x+30y≤600

x,y≥0

First we will convert the given inequations to equations. we get

300x+150y=7500

15x+30y=600

x=0 and y=0

Region represented by 300x+150y≤7500:

The line 300x+150y=7500 meets the coordinate axes at A(25, 0) and B(0, 50) respectively.

Clearly (0, 0) satisfies the equation 300x+150y=7500.

So the region which contains origin that represents the solution set of this inequation.

Region represented by 15x+30y≤600:

The line 15x+30y=600 meets the coordinate axes at C(40, 0) and D(0, 20) respectively.

Clearly (0, 0) satisfies the equation 15x+30y=600.

So the region in XY plane that contains the origin represents the solution set of this inequation.

Region represented by x,y≥0:

Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are O(0, 0), A(25, 0), D(0, 20) and E(20,10).

The value of objective function at these points are given in the following table.

Corner Pointsz=x+y
O(0, 0)0
A(25, 0)25
E(20, 10)30
D(0, 20)20

Thus maximum number of cakes is 30 and that can be made from 20 of the first kind and 10 of other kind.

Question 10. Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60 per kg and food Q costs Rs 80 per kg. Food P contains 3 units per kg of vitamin A and 5 units per kg of vitamin B while food Q contains 4 units per kg of vitamin A and 2 units per kg of vitamin B. Determine the minimum cost of the mixture.

Solution:

Let x units of food P and y units of food Q are mixed together to make the mixture.

The cost of food P is Rs 60 per kg and that of food Q is Rs 80 per kg.

So, x kg of food P and y kg of food Q will cost Rs(60x+80y).

Since one kg of food P contains 3 units of vitamin A and one kg of food Q contains 4 units of vitamin A, therefore x kg of food P and y kg of food Q will contain (3x+4y) units of vitamin A. But the mixture should contain at least 8 units of vitamin A.

Thus, 3x+4y≥8

Similarly x kg of food P and y kg of food Q will contain 5x+2y units of vitamin B. But the mixture should contain at least 11 units of vitamin B.

Thus, 5x+2y≥11

Now the mathematical formulation of given linear programming problem is

Minimize z=60x+80y

subject to

3x+4y≥8

5x+2y≥11

x,y≥0

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are A(8/3, 0), B(2, 1/2), C(0, 11/2).

The value of objective function at these points are given in the following table.

Corner pointz=60x+80y
A160(Minimum)
B160(Minimum)
C440

The smallest value of z is 160 at A(8/3, 0) and B(2, 1/2).

It can be verified that the open half-plane represented by 60x+80y<160 has no common points with the feasible region.

So, the minimum cost of mixture is Rs 160 since the minimum value of z is 160.

Question 11. One kind of cake requires 200g of flour and 25 gms of fat and another kind of cake requires 100 gms of flour and 50 grams of fat. Find the maximum number of cakes that can be made from 5 kg of flour and 1 kg of fat. Assuming that there is no storage of the other ingredients used in making the cakes.

Solution:

Let the number of cakes of one kind and another kind be x and y respectively.

Therefore, total number of cakes produced are (x+y).

One kind of cake requires 200g of flour and another kind of cake requires 100 gms of flour. 

So, x cakes of one kind and y cakes of another kind requires (200x+100y) grams of flour. But cakes should contain at most 5 kg of flour.

Thus, 200x+100y≤5000

⇒  2x+y≤50

One kind of cake requires 25g of fat and another kind of cake requires 50 gms of fat.

So, x cakes of one kind and y cakes of another kind requires (25x+50y) grams of fat. But cakes should contain at most 1 kg of fat.

Thus, 25x+50y≤1000

⇒ x+2y≤40

Now the mathematical formulation of given linear programming problem is

Minimize z=x + y

subject to

2x+y≤50

x+2y≤40

x,y≥0

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are O(0, 0),  A(25, 0), B(20, 10), C(0, 20).

The value of objective function at these points are given in the following table.

Corner pointz=x+y
O0+0=0
A25+0=25
B20+10=30(Maximum)
C0+20=20

Thus the maximum value of z is 30 at B(20,10).

Hence the maximum number of cakes which can be made are 30.

Question 12. A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 grams) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires at least 240 units of calcium, at least 460 units of iron and atmost 300 units of cholesterol. How many packets of each food should be used to minimize the amount of vitamin A in the diet? What is the minimum of vitamin A?

Solution:

Let x packets of food P and y packets of food Q be used to make the diet.

Each packet of food P contains 6 units of vitamin A and each packet of food Q contains 3 units of vitamin A.

Therefore, x packets of food P and y packets of food Q contains (6x+3y) units of vitamin A.

Since each packet of food P contains 12 units of calcium and each packet of food Q contains 3 units of calcium.

Therefore, x packets of food P and y packets of food Q will contain (12x+3y) units of calcium. But the diet should contain at least 240 units of calcium.

Thus 12x+3y≥240

⇒ 4x+y≥80

Similarly,  x packets of food P and y packets of food Q contains (4x+20y) units of iron.

But the diet should contain at least 460 units of iron.

Thus 4x+20y≥460

⇒ x+5y≥115

Also, x packets of food P and y packets of food Q contains (6x+4y) units of cholesterol.

But the diet should contain atmost 300 units of cholesterol.

Thus 6x+4y≤300

⇒ 3x+2y≤150

Now the mathematical formulation of given linear programming problem is

Minimize z=6x+3y

subject to

4x+y≥80

x+5y≥115

3x+2y≤150

x,y≥0

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are A(2, 72), B(15, 20), C(40, 15).

The value of objective function at these points are given in the following table.

Corner Pointz=6x+3y
A6(2)+3(72)=228
B6(15)+3(20)=150
C6(40)+3(15)=285

The smallest value of z is 150 which is obtained at point B(15, 20).

Thus,15 packets of food P and 20 packets of food Q should be used to minimize the amount of vitamin A on the diet.

Hence, the minimum amount of vitamin A is 150 units.

Question 13. A farmer mixes two brands of P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units,45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a cost per bag? What is the minimum cost of mixture per bag?

Solution:

Let x bags of brand P and y bags of brand Q should be mixed to produce the mixture.

Each bag of brand P costs Rs 250 per bag and each bag of brand Q costs Rs 200 per bag.

Therefore, x bags of brand P and y bags of brand Q costs Rs(250x+200y).

Since each bag of brand P contains 3 units of nutritional element A and each bag of brand Q contains 1.5 units of nutritional element A.

Therefore, x bags of brand P and y bags of brand Q will contain (3x+1.5y) units of nutritional element A.

But the minimum requirement of nutrients A is 18 units.

Thus, 3x+1.5y≥18

⇒  2x+y≥12

Similarly, x bags of brand P and y bags of brand Q will contain (2.5x+11.25y) units of nutritional element B.

But the minimum requirement of nutrients B is 45 units.

Thus, 2.5x+11.25y≥45

⇒  2x+9y≥36

Also, x bags of brand P and y bags of brand Q will contain (2x+3y) units of nutritional element C.

But the minimum requirement of nutrients C is 24 units.

Thus, 2x+3y≥24

Now the mathematical formulation of given linear programming problem is

Minimize z=250x+200y

subject to

 2x+y≥12

2x+9y≥36

2x+3y≥24

x,y≥0

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are A(18, 0), B(9, 2), C(3, 6), D(0, 12).

The value of objective function at these points are given in the following table.

Corner pointz=250x+200y
A250(18)+200(0)=4500
B250(9)+200(2)=2650
C250(3)+200(6)=1950
D250(0)+200(12)=2400

The smallest value of z is 1950 obtained at C(3, 6).

So, 3 bags of brand P and 6 bags of brand Q should be used to minimize the cost.

Hence, the minimum cost of mixture per bag is Rs 1950.

Question 14. A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B, and 8 units of vitamin C. The vitamin contents of one kg food are given below:

FoodVitaminAVitaminBVitaminC
X123
Y221

One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of mixture which will produce the required diet?

Solution:

Let x kg of food X and y kg of food Y are mixed together to make the mixture.

The cost of food X is Rs 16 per kg and that of food Y is Rs 20 per kg.

So, x kg of food X and y kg of food Y will cost Rs (16x+20y).

Since one kg of food X contains 1 unit of vitamin A and one kg of food Y contains 2 units of vitamin A. Therefore, x kg of food X and y kg of food Y will contain (x+2y) units of vitamin A.

But the mixture should contain at least 10 units of vitamin A.

Thus, x+2y≥10

Similarly x kg of food X and y kg of food Y will contain (2x+2y) units of vitamin B. But the mixture should contain at least 12 units of vitamin B.

Thus, 2x+2y≥12

⇒ x+y≥6

Also, x kg of food X and y kg of food Y will contain (3x+y) units of vitamin C. But the mixture should contain at least 8 units of vitamin C.

Thus, 3x+y≥8

Now the mathematical formulation of given linear programming problem is

Minimize z=16x+20y

subject to

x+2y≥10

x+y≥6

3x+y≥8

x,y≥0

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are A(10, 0), H(2, 4), G(1, 5), F(0,8).

The value of objective function at these points are given in the following table.

Corner Pointz=16x+20y
A16(10)+20(0)=160
H16(2)+20(4)=112(Minimum)
G16(1)+20(5)=116
F16(10)+20(8)=160

The smallest value of z is 112 obtained at H(2, 4).

Hence the least cost of mixture which will produce the required diet is Rs 112.

Question 15. A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid at least 270 kg of potash and almost 310 kg of chlorine.

Kg per bagBrand PBrand Q
Nitrogen33.5
Phosphoric Acid12
Potash31.5
Chlorine1.52

If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added to the garden?

Solution:

Let x bags of fertilizer of brand P and y bags of fertilizer of brand Q are used in the garden.

One bag of brand P contains 3 kg of nitrogen and one bag of brand Q contains 3.5 kg of nitrogen, therefore, x bags of brand P and y bags of brand Q will contain (3x+3.5y) kg of nitrogen.

Since, One bag of brand P contains 1 kg of phosphoric acid and one bag of brand Q contains 2 kg of phosphoric acid, therefore, x bags of brand P and y bags of brand Q will contain (x+2y) kg of phosphoric acid.

But the garden needs at least 240 kg of phosphoric acid.

Thus, x+2y≥240

Similarly, One bag of brand P contains 3 kg of potash and one bag of brand Q contains 1.5 kg of potash, therefore, x bags of brand P and y bags of brand Q will contain (3x+1.5y) kg of potash.

But the garden needs at least 270 kg of  potash.

Thus, 3x+1.5y≥270

⇒  2x+y≥180

Also, One bag of brand P contains 1.5 kg of Chlorine and one bag of brand Q contains 2 kg of Chlorine, therefore, x bags of brand P and y bags of brand Q will contain (1.5x+2y) kg of Chlorine.

But the garden needs atleast 310 kg of  Chlorine.

Thus, 1.5x+2y≤310

Now the mathematical formulation of given linear programming problem is

Minimize z=3x+3.5y

subject to

x+2y≥240

2x+y≥180

1.5x+2y≤310

x,y≥0

The feasible region determined by the given constraints can be graphically represented as

The corner points of a Feasible region are A(40, 100), B(140, 50), C(20, 140).

The value of objective function at these points are given in the following table.

Corner pointz=3x+3.5y
A(40, 100)3(40)+3.5(100)=470
B(140, 50)3(140)+3.5(50)=595
C(20, 140)3(20)+3.5(140)=550

The smallest value of z is 470 obtained at A(40, 100).

Hence 40 bags of fertilizer of brand P and 100 bags of fertilizer of brand Q are used in the garden. The minimum amount of nitrogen added in the garden is 470 kg.

Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.4 | Set 1

  • Last Updated : 26 May, 2021

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Question 1. If a young man drives his vehicle at 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs 5/per km. He has Rs 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.

Solution:

Assume the young man drives x km at a speed of 25 km/hr and y km at a speed of 40 km/hr. 

Thus, x, y ≥ 0

Given: If he spends Rs 2/km if he drives at a speed of 25 km/hr 

Rs 5/km if he drives at a speed of 40 km/hr. 

Hence, 

Money spent by him when he traveled x km and y km will be Rs 2x and Rs 5y respectively.

Thus, here it is given that he can spend Rs100 maximum.

Therefore, 

2x + 5y ≤ 100

So, for traveling with a speed of 25 km/hr the time that he will be spending = x/25 hr 

So, for traveling with a speed of 40km/hr the time that he will be spending = y/40 hr 

And also, 

The time he is having is 1 hour.

x/25 + y/40 ≤ 1

40x + 25y ≤ 1000

The distance covered Z = x + y 

That is to be maximized.

Therefore, the formula of the given linear programming problem is 

Max Z = x + y 

Subject to constraints

2x + 5y ≤ 100

40x + 25y ≤ 1000

x, y ≥ 0

First we will convert in-equations as follows:

2x + 5y = 100

40x + 25y = 1000

x = 0 and y = 0.

The area shown by 2x + 5y ≤ 100

The line 2x + 5y = 100 connects the coordinate axes at A(50, 0) and B(0, 20) respectively. 

After connecting these points, we will get the line 2x + 5y = 100. 

Thus, (0, 0) assure the 2x + 5y = 100. 

As, the area that consists the origin shows the solution set of the in-equation 2x + 5y ≤ 100

The area shown by 40x + 25y ≤ 1000

The line 40x + 25y = 1000 connects the coordinate axes at C(25, 0) and D(0, 40) respectively. 

After connecting these points, we will get the line 2x + y = 12. 

Thus, (0, 0) assure the 40x + 25y = 1000. 

As, the area that consists of the origin shows the solution set of the in-equation 40x + 25y ≤ 1000

The area shown by x ≥ 0, 

y ≥ 0 :

Since, all the points in the first quadrant assure these in-equations. 

As, the first quadrant is the area shown by the in-equations x ≥ 0 

and 

y ≥ 0.

The suitable area is determined by the system of constraints

2x + 5y ≤ 100, 

40x + 25y ≤ 1000, 

x ≥ 0 

and 

y ≥ 0 are as follows

The corner points are O(0, 0), B(0, 20), E(50/3, 40/3), and C(25, 0). The value of Z at these corner points are as follows:

Corner PointsZ = x + y
(0, 0)0
(0, 20)20
50/3, 40/330
(25, 0)25

The maximum value of Z is 30 that is acquired at E.

Therefore, 

The maximum distance traveled by the young man is 30 kms, if he drives 50/3 km at a speed of 25 km/hr and 40/3 km at a speed of 40 km/hr.

Question 2. A manufacturer has three machines installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:

ItemNumber of hours required by the machine
 IIIIII
A121
B21

He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B. Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.

Solution:

Assume x units of item A and y units of item B be produced. 

Therefore, x, y ≥ 0.

Given:

\frac{5}{4}

Machines I and II are efficient of being operated for at most 12 hours whereas Machine III should operate at least for 5 hours a day.

As per in the question, 

The constraints are

x + 2y ≤ 12

2x + y ≤ 12

x + 5/4

y ≥ 5

He makes a profit of Rs 6.00 on item A and Rs. 4.00 on item B. Profit made by him in manufacturing x items of A and y items of B is 6x + 4y.

Total profit is Z 

Z = 6x + 4y that is to be maximized

Therefore, the formula of the given linear programming problem is 

Max Z = 6x + 4y, 

Subject to constraints

x + 2y ≤ 12

2x + y ≤ 12

x + 5/4

y ≥ 5x, 

y ≥ 0

First, we have to convert the in-equations into equations as follows:

x + 2y = 12, 

2x + y = 12, 

x + 5/4

y = 5, 

x = 0 and 

y = 0.

The area shown by x + 2y ≤ 12

The line x + 2y = 12 connects the coordinate axes at A(12, 0) and B(0, 6) respectively. 

After connecting these points, we will get the line x + y = 12. 

As (0, 0) assure the x + 2y = 12. 

As, the area that consists the origin shows the solution set of the in-equation x + 2y ≤ 12

The area shown by 2x + y ≤ 12

The line 2x + y = 12 connects the coordinate axes at C(6, 0) and D(0, 12) respectively. 

After connecting these points, we will connect the line 2x + y = 12. 

As (0, 0) assure the 2x + y = 12. 

As, the area that consists the origin shown the solution set of the in-equation 2x + y ≤ 12

The area shown by x + 5/4y ≥ 5

The line x + 5/4y ≥ 5 connects the coordinate axes at E(5, 0) and F(0, 4) respectively. 

After connecting these points, 

We will get the line x + 5/4y = 5. 

As (0, 0) assure the x + 5/4y ≥ 5. 

As, the area that does not consist the origin shows the solution set of the in-equation x + 5/4y ≥ 5

The area shown by x ≥ 0, 

y ≥ 0 :

As all the points in the first quadrant assure these in-equations. 

Thus, the first quadrant is the area shown by the in-equations x ≥ 0 

and 

y ≥ 0.

The suitable area determined by the system of constraints

x + 2y ≤ 12, 

2x + y ≤ 12, 

x + 5/4y ≥ 5, x, 

y ≥ 0 are as follows.

Therefore, the maximum profit is of Rs 40 gained when 4 units each of item A and B are manufactured.

The corner points are D(0, 6), I(4, 4), C(6, 0), G(5, 0), and H(0, 4). 

The values of Z at these corner points are as follows:

Corner Points Z = 6x + 4y
D24
I40
C36
G30
H16

Hence, the maximum value of Z is 40 which is acquired at I(4, 4).

Question 3. Two tailors, A and B earn Rs 15 and Rs 20 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce atleast 60 shirts and 32 pants at a minimum labour cost?

Solution:

Assume thet tailor A work for x days and tailor B work for y days.

In a day; A can stitch 6 shirts and 4 pants whereas B can stitch 10 shirts and 4 pants.

Therefore, in x days; A can stitch 6x shirts and 4x pants whereas in y days B can stitch 10y shirts and 4y pants.

Given: The minimum need of the shirt and pants are respectively 60 and 32.

Therefore,

6x + 10y ≥ 60

4x + 4y ≥ 32

So further, it is given that A and B earn Rs 15 and Rs 20 per day respectively. 

Therefore, A earn Rs 15x and B earns Rs 20y.

Assume the total cost is Z

Z = 15x + 20y

As days cannot be negative.

x, y ≥ 0 (always)

Min Z = 15x + 20y

Subject to constraints

6x + 10y ≥ 60

4x + 4y ≥ 32

x, y ≥ 0

First we will convert in-equations into equations as follows:

6x + 10y = 60, 

4x + 4y = 32, 

x = 0, 

y = 0

The area shown by 6x + 10y ≥ 60

The line 6x + 10y = 60 connects the coordinate axes at A(10,0) and B(0,6) respectively. 

After connecting these points we will get the line 6x + 10y = 60. 

As, (0, 0) assure the 6x + 10y ≥ 60. 

Since, the area that does not consist the origin shown the solution set of the in-equation 6x + 10y ≥ 60

Area shown by 4x + 4y ≥ 32

The line 4x + 4y = 32 connects the coordinate axes at C(8,0) and D(0,8) respectively. 

After connecting these points we will get the line 4x + 4y = 32. 

As, (0, 0) assure the 4x + 4y ≥ 32. 

As, the area that does not consist the origin shows the solution set of the in-equation 4x + 4y ≥ 32.

The area shown by x ≥ 0, 

y ≥ 0 :

Since, all the points in the first quadrant assure these in-equations. 

As, the first quadrant is the area shown by the in-equations x ≥ 0 

and 

y ≥ 0.

The suitable area determined by the system of constraints 6x + 10y ≥ 60, 

4x + 4y ≥ 32

x, y ≥ 0 are as follows.

The corner points are D(0, 8), E(5, 3), A(10, 0). 

The values of Z at these corner points are as follows:

Corner PointsZ = 15x + 20y
D160
E135
A150

The minimum value of Z is 135 which is acquired at E(5, 3).

Therefore, For minimum labor cost, A must work for 5 days and B must work for 3 days.

Question 4. A factory manufactures two types of screws, A and B, each type requiring the use of two machines – an automatic and a hand-operated. It takes 4 minute on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws ‘A’, while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws ‘A’ at a profit of 70 P and screws ‘B’ at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

Solution:

Assume the factory manufacture x screws of type A and y screws of type B on each day,

Thus, 

x, y ≥  0.

Given information in a table as;

 Score AScore BScore C
Automatic Machine (min)464 × 60 = 240
Hand Operated Machine (min)634 × 60 = 240

4x + 6y ≤ 240

6x + 3y ≤ 240

The manufacturer can sell a package of screws ‘A’ at a profit of Rs 0.7 and screws ‘B’ at a profit of Re 1.

Assume the total profit is Z, 

Z = 0.7x + 1y

Therefore, the formula of the given linear programming problem is 

Maximize Z = 0.7x + 1y

Subject to the constraints,

4x + 6y ≤ 240

6x + 3y ≤ 240

x, y ≥ 0

First, we will convert the in-equations into equations as follows:

4x + 6y = 240, 6x + 3y = 240, x = 0, y = 0.

Area shown by 4x + 6y ≥  240

The line 4x + 6y = 240 connects the coordinate axes at A(60, 0) and B(0, 40) respectively.

After connecting these points we will get the line 4x + 6y = 240. 

As, (0, 0) assure the 4x + 6y ≥  240. 

As, the area that consists the origin showss the solution set of the in-equation 4x + 6y ≥  240.

The area shown by 6x + 3y ≥  240

The line 6x + 3y = 240 connects the coordinate axes at C(40, 0) and d(0, 80) respectively. 

After connecting these points we will get the line 6x + 3y = 240. 

As, (0, 0) assure the 6x + 3y ≥  240. 

Since, the area that consist the origin shows the solution set of the in-equation 6x + 3y ≥  240.

The area shown by 

x ≥  0, 

y ≥ 0 :

As, all the points in the first quadrant assure these in-equations. 

Since, the first quadrant is the area shown by the in-equations 

x ≥  0 

and 

y ≥  0.

The suitable are determined by the system of constraints 

4x + 6y ≤ 240, 

6x + 3y ≤ 240, 

x ≥  0,

y ≥  0 are as follows.

The corner points are C(40, 0), E(30, 20), B(0, 40). The values of Z at these corner points are as follows

Corner PointZ = 7x + 10y
C(40, 0)280
E(30, 20)410
B(0, 40)400

The maximum value of Z is 410 at (30, 20).

Therefore, the factory must manufacture 30 packages of screws A and 20 packages 

of screws b to get the maximum profit of Rs 410.

Question 5. A company produces two types of leather belts, say type A and B. Belt A is a superior quality and belt B is of a lower quality. Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type B, only 700 buckles are available per day. How should the company manufacture the two types of belts in order to have a maximum overall profit?

Solution:

Assume the company manufactures x belts of types A and y belts of type B. 

Number of belts cannot be negative. 

Thus, 

x, y ≥ 0 (always)

It is given that leather is enough only for 800 belts per day (both A and B combined).

Thus,

x + y ≤ 800

It is given that the rate of manufacture of belts of type B is 1000 per day. 

Therefore, the time taken to produce y belts of type B is y/1000.

And, 

As each belt of type A needs twice as much time as a belt of type B, the rate of production of belts of type A is 500 per day and thus, total time taken to produce x belts of type A is x/500 

Therefore, 

We have,

x/500 + y/1000 ≤ 1

Or, 

2x + y ≤ 1000

Belt A needs fancy buckle and only 400 fancy buckles are accessible for this per day.

x ≤ 400

For Belt of type B only 700 buckles are accessible per day.

y ≤ 700

Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. 

Thus, profit gained on x belts of type A and y belts of type B is Rs 2x and Rs 1.50y respectively. 

Therefore, the total profit would be Rs(2x + 1.50y). 

Assume Z denote the total profit

Z = 2x + 1.50y

Therefore, the mathematical formulation of the given linear programming problem is;

Max Z = 2x + 1.50y subject to

x + y ≤ 800

2x + y ≤ 1000

x ≤ 400

y ≤ 700

First we will convert these in-equations into equations as follows:

x + y = 800

2x + y = 1000

x = 400

y = 700

The area shown by x + y = 800

The line x + y = 800 meets the coordinate axes at A(800, 0) and B(0, 800) respectively. 

After connecting these points we obtain the line x + y = 800. 

As (0, 0) assure the x + y ≥ 800. 

As, the area that consist the origin shows the solution set of the in-equation x + y ≥ 800.

The area shown by 2x + y ≥  1000

The line 2x + y = 1000 connects the coordinate axes at C(500, 0) and D(0, 1000) respectively. 

After connecting these points we will get the line 2x + y = 1000. 

As, (0, 0) assure the 2x + y ≥ 1000. 

As, the area that consists the origin shows the solution set of the in-equation 2x + y ≥ 1000.

The area shown by x ≤ 400

The line x = 400 will pass through (400, 0). 

The area to the left of the line x = 400 will assure the in-equation x ≤ 400

The area shown by y ≤ 700

The line y = 700 will pass through (0, 700). 

The area to the left of the line y = 700 will assure the in-equation y ≤ 700.

The area shown by x ≥ 0, 

y ≥ 0 :

As, all the points in the first quadrant assure these in-equations. 

As, the first quadrant is the area shown by the in-equations x ≥ 0 

and y ≥ 0.

The suitable area determined by the system of constraints x + y ≤ 800, 

2x + y ≤ 1000, 

x ≤ 400,

y ≤ 700

The corner points are F(0, 700), G(200, 600), H(400, 200), E(400, 0). 

The values of Z at these corner points are as follows

Corner PointsZ = 2x + 1.5y
F(0, 700)1050
G(200, 600)1300
H(400, 200)1100
E(400, 0)800

The maximum value of Z is 1300 which is attained at G(200, 600).

Therefore, the maximum profit acquires is Rs 1300 when 200 belts of type A and 600 belts of type B are produced.

Question 6. A small manufacturer has employed 5 skilled men and 10 semi-skilled men and makes an article in two qualities deluxe model and anordinary model. The making of a deluxe model requires 2 hrs. work by a skilled man and 2 hrs. work by a semi-skilled man. The ordinarymodel requires 1 hr by a skilled man and 3 hrs. by a semi-skilled man. By union rules no man may work more than 8 hrs per day. The manufacturers clear profit on deluxe model is Rs 15 and on an ordinary model is Rs 10. How many of each type should be made in order to maximize his total daily profit.

Solution:

Assume x articles of deluxe model and y articles of an ordinary model be made.

Numbers cannot be negative.

Therefore,

x, y ≥ 0 (alwaya)

As per the question, 

The profit on each model of deluxe and ordinary type model are Rs 15 and Rs 10.

Thus, profits on x deluxe model and y ordinary models are 15x and 10y.

Assume Z be total profit,

So,

Z = 15x + 10y

As, the making of a deluxe and ordinary model needs 2 hrs. and 1 hr work by skilled men, 

so, x deluxe and y ordinary models needs 2x and y hours of skilled men but time needed 

by skilled men is 5 × 8 = 40 hours.

Thus,

2x + y ≤ 40 {First Constraint}

As, the making of a deluxe and ordinary model needs 2 hrs. and 3 hrs work by semi-skilled men, 

Since, x deluxe and y ordinary models need 2x and 3y hours of skilled men but time 

need by skilled men is 10 × 8 = 80 hours.

As,

2x + 3y ≤ 80 {Second constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 15x + 10y

Subject to constraints,

2x + y ≤ 40

2x + 3y ≤ 80

x, y ≥ 0

Area 2x + y ≤ 40: 

Line 2x + 4y = 40 connects axes at 

A1(20, 0), B1(0, 40) respectively. 

The area having origin shows 2x + 3y ≤ 40 as (0, 0) assure 2x + y ≤ 40

Area 2x + 3y ≤ 80:

Line 2x + 3y = 80 connects axes at 

A2(40, 0), B2(0, 80/3) respectively. 

The area having origin shows 2x + 3y ≤ 80.

The corner points are A1(20, 0), P(10, 20), B2(0, 80/3).

The value of Z = 15x + 10y at these corner points are

Corner PointsZ = 15x + 10y
A1300
P350
B2800/3

The maximum value of Z is 350 which is acquired at P(10, 20).

Therefore, maximum profit is attained when 10 units of deluxe model and 20 units of ordinary model is produced.

Question 7. A manufacturer makes two types A and B of tea-cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:

 Machines
 IIIIII
A12186
B609

Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea-cups of type A and 30 of type B should be manufactured in a day to get the maximum profit.

Solution:

Assume the needed number of tea cups of Type A and B are x and y respectively.

As, the profit on each cup A is 75 paise and that on each cup B is 50 paise. 

So, the profit on x tea cup of type A and y tea cup of type B are 75x and 50y respectively.

Assume total profit on tea cups be Z, so

Z = 75x + 50y

Since, every tea cup of type A and B needs to work machine I for 12 and 6 minutes respectively. 

So, x tea cups of Type A and y tea cups of Type B needs to work on machine I for 12x and 6y minutes.

Total time available on machine I is 6 × 60 = 360 minutes. 

As

12x + 6y ≤ 360 {First Constraint}

Thus, every tea cup of type A and B need to work machine II for 18 and 0 minutes respectively. 

So, x tea cups of Type A and y tea cups of Type B need to work on machine IIII for

18x and 0y minutes respectively.

Total time available on machine I is 6 × 60 = 360 minutes. 

So,

18x + 0y ≥ 360

x ≤ 20 {Second Constraint}

As, every tea cup of type A and B need to work machine III for 6 and 9 minutes respectively 

So, x tea cups of Type A and y tea cups of Type B need to work on machine I 

for 6x and 9y minutes respectively.

Total time available on machine I is 6 × 60 = 360 minutes. 

So,

6x + 9y ≤ 360 {Third Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 75x + 50y

Subject to constraints,

12x + 6y ≤ 360

x ≤ 20

6x + 9y ≤ 360

x, y ≥ 0 {Since production of tea cups can not be less than zero}

Area 12x + 6y ≤ 360: line 12x + 6y = 360 connects axes at A(30, 0), B(0, 60) 

The area having origin shows 12x + 6y ≤ 360 as (0, 0) assure 12x + 6y ≤ 360

Area x ≤ 20: line x = 20 is parallel to y-axis and connects x-axes at C(20, 0). 

The area having origin shows x ≤ 20

As (0, 0) assure x ≤ 20.

Area 6x + 9y ≤ 360: line 6x + 9y = 360 connects axes at E(60, 0), F(0, 40) respectively. 

The area having origin shows 6x + 9y ≤ 360 as (0, 0) assure 6x + 9y ≤ 360.

Area x, y ≥ 0: it shows the first quadrant.

The shaded area is the suitable area determined by the constraints,

12x + 6y ≤ 360

x ≤ 20

6x + 9y ≤ 360

x, y ≥ 0

The corner points are F(0, 40), G(15, 30), H(20, 20), C(20, 0).

The values of Z at these corner points are as follows

Corner PointsZ = 75x + 50y
F2000
G2624
H2500
C1500

Here Z is maximum at G(15, 30).

Hence, 15 teacups of Type A and 30 tea cups of Type B are required to maximize the profit.

Question 8. A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:

 Area occupied by the machineLabour force for each machineDaily output in units
Machine A1000 sq. m12 men60
Machine B 1200 sq. m8 men40

He has an area of 7600 sq. m available and 72 skilled men who can operate the machines. How many machines of each type should he buy to maximize the daily output?

Solution:

Assume needed number of machine A and B are x and y respectively.

Thus, products of each machine A and B are 60 and 40 units daily respectively. 

As, production by x number of machine A and y number of machine B are 60x and 40y respectively.

Assume Z is total output daily, 

So,

Z = 60x + 40y

As, every machine of type A and B needs 1000sq. m and 1200 s. m area so, 

x machine of type A and y machine of type B require 1000x and 1200y sq. m area but,

Total available area for machine is 7600 sq. m. 

So,

1000x + 1200y ≤ 7600

or, 

5x + 6y ≤ 38. {First Constraint}

As, every machine of type A and B needs 12 men and 8 men to work respectively. 

So, x machine of type A and y machine of type B need 12x and 8y men to work respectively.

But total men available for work are 72.

So,

12x + 8y ≤ 72

3x + 2y ≤ 18 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 50x + 40y

Subject to constraints,

5x + 6y ≤ 38

3x + 2y ≤ 18

x, y ≥ 0 {Since number of machines can not be less than zero}

Area 5x + 6y ≤ 38: line 5x + 6y = 38 connects the axes at A(38/5, 0), B(0, 19/3) respectively.

Area having the origin shows 5x + 6y ≤ 38 as origin satisfies 5x + 6y ≤ 38

Area 3x + 2y ≤ 18: line 3x + 2y = 18 connects the axes at C(6, 0), D(0, 9) respectively.

Area having the origin shows 3x + 2y ≤ 18 as origin assure 3x + 2y ≤ 18.

Area x, y ≥ 0: it represents the first quadrant.

Shaded region shows the suitable area.

The corner points are O(0, 0), B(0, 19/3), E(4, 3), C(6, 0).

Therefore, the values of Z at these corner points are as follows:

Corner PointsZ = 60x + 40y
O0
B253.3
E360
C360

The maximum value of Z is 360 which is acquired at E(4, 3), C(6, 0).

Therefore, the maximum output is Rs 360 archive when 4 units of type A and 

3 units of type B or 6 units of type A and 0 units of type B are manufactured.

Question 9. A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 gm of silver and 1 gm of gold while that of type B requires 1 gm of silver and 2 gm of gold. The company can produce 9 gm of silver and 8 gm of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?

Solution:

Assume needed number of goods A and B are x and y respectively.

As, profits of each A and B are Rs. 40 and Rs. 50 respectively. 

Thus, profits on x number of type A and y number of type B are 40x and 50y respectively.

Assume Z is total output daily, 

So,

Z = 40x + 50y

So, every A and B needs 3 grams and 1 gram of silver respectively. 

Thus, x of type A and y of type B need 3x and y of silver respectively. 

But,

Total silver available is 9 grams. So,

3x + y 

9 {First Constraint}

So, every A and B needs 1 gram and 2 grams of gold respectively. 

Thus, x of type A and y of type B require x and 2y respectively.

But total gold available is 8 grams.

Thus,

x + 2y ≤ 8 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 40x + 50y

Subject to constraints,

3x + y ≤ 9

x + 2y ≤ 8

x, y ≥ 0 {Since production of A and B can not be less than zero}

Area 3x + y ≤ 9: line 3x + y = 9 connects the axes at A(3, 0), B(0, 9) respectively.

The area having the origin shows 3x + y ≤ 9

As, origin assure 3x + y ≤ 9.

Area x + 2y ≤ 8: line x + 2y = 8 connects the axes at C(8, 0), D(0, 4) respectively.

Area having the origin shows x + 2y ≤ 8 as origin assure x + 2y ≤ 8.

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), D(0, 4), E(2, 3), A(3, 0)

The values of Z at these corner points are as follows

Corner PointsZ = 40x + 50y
O0
D200
E230
A120

The maximum value of Z is 230 which is obtained at E(2, 3).

Therefore, the maximum profit is of Rs 230 when 2 units of Type A 3 units of Type B are produced.

Question 10. A manufacturer of Furniture makes two products: chairs and tables. Processing of these products is done on two machines A and B. A chair requires 2 hrs on machine A and 6 hrs on machine B. A table requires 4 hrs on machine A and 2 hrs on machine B. There are 16 hrs 0of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his profit.

Solution:

Assume daily production of chairs and tables be x and y respectively.

So, profits of every chair and table is Rs. 3 and Rs. 5 respectively. 

Thus, profits on x number of type A and y number of type B are 3x and 5y respectively.

Assume Z is total output daily, 

So,

Z = 3x + 5y

As, every chair and table needs 2 hrs and 3 hrs on machine A respectively. 

So, x number of chair and y number of table need 2x and 4y hrs on machine A respectively. 

But,

Total time available on Machine A is 16 hours. 

So,

2x + 3y ≤ 16

x + 2y ≤ 8 {First Constraint}

So, every chair and table needs 6 hrs and 2 hrs on machine B respectively. 

Since, x number of chair and y number of table need 6x and 2y hrs on machine B respectively. 

But,

Total time need on Machine B is 30 hours. So,

6x + 2y ≤ 30

3x + y ≤ 15 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 3x + 5y

Subject to constraints,

x + 2y ≤ 8

3x + y ≤ 15

x, y ≥ 0 {Since production of chairs and tables can not be less than zero}

Area x + 2y ≤ 8: line x + 2y = 8 connects the axes at A(8, 0), B(0, 4) respectively.

Area having the origin shows x + 2y ≤ 8

As, origin assure x + 2y ≤ 8.

Area 3x + y ≤ 15: line 3x + y = 15 connects the axes at C(5, 0), D(0, 15) respectively.

Area having the origin shows 3x + y ≤ 15 as origin assure 3x + y ≤ 15

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), B(0, 4), E(22/5, 9/5), and C(5, 0).

The values of Z at these corner points are as follows,

Corner PointsZ = 3x + 5y
O0
B20
E22.2
C15

The maximum value of Z is 22.2 which is obtained at E(22/5, 9/5).

Therefore, the maximum profit of Rs 22.2 when 22/5 units of chair and 9/5 units of table are produced.

Question 11. A furniture manufacturing company plans to make two products: chairs and tables. From its available resources which consists of 400 square feet to teak wood and 450 man-hours. It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yield sa profit of Rs 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs 80. How many items of each product should be produced by the company so that the profit is maximum?

Solution:

Assume needed production of chairs and tables be x and y respectively.

As, profits of every chair and table is Rs. 45 and Rs. 80 respectively. 

Thus, profits on x number of type A and y number of type B are 45x and 80y respectively.

Assume Z is total output daily, 

So, Z = 45x + 80y

As, every chair and table needs 5 sq. ft and 80 sq. ft of wood respectively. 

As, x number of chair and y number of table need 5x and 80y sq. ft of wood respectively. 

But, 400 sq. ft of wood is accessible. 

So, 5x + 80y ≤ 400

x + 4y ≤ 80 {First Constraint}

As, every chair and table needs 10 and 25 men – hours respectively. 

As, x number of chair and y number of table need 10x and 25y men – hours respectively. 

But, 

Only 450 hours are available . 

So,

10x + 25y ≤ 450

2x + 5y ≤ 90 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 45x + 80y

Subject to constraints,

x + 4y ≤ 80

2x + 5y ≤ 90

x, y ≥ 0 {Since production of chairs and tables can not be less than zero}

Area x + 4y ≤ 80: line x + 4y = 80 connects the axes at A(80, 0), B(0, 20) respectively.

Area having the origin shows x + 4y ≤ 80 as origin assure x + 4y ≤ 80

Area 2x + 5y ≤ 90: line 2x + 5y = 90 connects the axes at C(45, 0), D(0, 20) respectively.

Area having the origin shows 2x + 5y ≤ 90

As origin assure 2x + 5y ≤ 90

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), D(0, 18), C(45, 0).

The values of Z at these corner points are as follows:

Corner PointsZ = 45x + 80y
O0
D1440
C2025

The maximum value of Z is 2025 which is obtained at C(45, 0).

Therefore, the maximum profit of Rs 2025 is attained when 45 units of chairs and no units of tables are produced.

Question 12. A firm manufactures two products A and B. Each product is processed on two machines M1 and M2. Product A requires 4 minutes of processing time on M1 and 8 min. on M2 ; product B requires 4 minutes on M1 and 4 min. on M2. The machine M1 is available for not more than 8 hrs 20 min. while machine M2 is available for 10 hrs. during any working day. The products A and B are sold at a profit of Rs 3 and Rs 4 respectively. Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit and solve it graphically.

Solution:

Assume needed production of product A and B be x and y respectively.

As, profit on every product A and B are Rs. 3 and Rs. 4 respectively. 

Thus, profits on x number of type A and y number of type B are 3x and 4y respectively.

Assume Z is total output daily, 

So,

Z = 3x + 4y

Thus, every A and B needs 4 minutes each on machine M1

As, x of type A and y of type B need 4x and 4y minutes respectively. 

But,

Total time accessible on machine M1 is 8 hours 20 minutes = 500 minutes.

Thus,

4x + 4y ≤ 500

x + y ≤ 125 {First Constraint}

So, every A and B needs 8 minutes and 4 minutes on machine Mrespectively. 

Thus, x of type A and y of type B need 8x and 4y minutes respectively. 

But,

Total time accessible on machine M1 is 10 hours = 600 minutes.

Thus,

8x + 4y ≤ 600

2x + y ≤ 150 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 3x + 4y

Subject to constraints,

x + y ≤ 125

2x + y ≤ 150

x, y ≥ 0 {Since production of A and B can not be less than zero}

Area x + y ≤ 125: line x + y = 125 connects the axes at A(125, 0), B(0, 125) respectively.

Area having the origin shows x + y ≤ 125 as origin assure x + y ≤ 125.

Area 2x + y ≤ 150: line 2x + y = 150 connects the axes at C(75, 0), D(0, 150) respectively.

Area having the origin shows 2x + y ≤ 150 as origin assure 2x + y ≤ 150.

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), B(0, 125), E(25, 100), and C(75, 0).

The values of Z at these corner points are as follows:

Corner PointsZ = 3x + 4y
O0
B500
E475
C225

The maximum value of Z is 500 which is obtained at B(0, 125).

Therefore, the maximum profit is Rs 500 attained when no units of product A and 

125 units of product B are manufactured.

Question 13. A firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programming problem for maximum profit and solve it graphically.

Solution:

Assume x units of item A and y units of item B were manufactured.

Numbers of items cannot be negative. 

Therefore,

x, y ≥ 0 (always)

Given: The information can be tabulated as follows:

Products MotorsTransformers
A(x)34
B(y)24
Availability210300

Thus, after this it is given that type B is an export model, whose supply is restricted to 65 units per month.

So, 

The constraints are

3x + 2y ≤ 210

4x + 4y ≤ 300

y ≤ 65

A and B can make profit of Rs 20 and Rs 30 per unit respectively.

Hence, profit gained from x units of item A and y units of item B is Rs 20x and 30y respectively.

Assume total profit be Z,

So,

Total Profit = Z = 20x + 30y as per the question is to be maximised.

Therefore, the formula of the given linear programming problem is 

Max Z = 20x + 30y

Subject to constraints

3x + 2y ≤ 210

4x + 4y ≤ 300

y ≤ 65

x, y ≥ 0

Area shown by 3x + 2y ≤ 210: The line 3x + 2y = 210 connects the axes at A(70, 0), B(0, 105) respectively.

Area having the origin shows 3x + 2y ≤ 210 as origin assure 3x + 2y ≤ 210.

Area shown by 4x + 4y ≤ 300: The line 4x + 4y = 300 connects the axes at C(75, 0), D(0, 75) respectively.

Area having the origin shows 4x + 4y ≤ 300 as origin assure 4x + 4y ≤ 300

y = 65 is the line passing through the point E(0, 65) and is parallel to x – axis.

Area x,y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), E(0, 65), G(10, 65), F(60, 15) and A(70, 0).

The values of Z at these corner points are as follows:

Corner PointsZ = 20x + 30y
O0
E1950
G2150
F1650
A1400

The maximum value of Z is 2150 which is obtained at G(10, 65).

Therefore, the maximum profit is Rs. 2150 attained when 10 units of item A and 

65 units of item B are manufactured.

Question 14. A factory uses three different resources for the manufacture of two different products, 20 units of resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2, and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured to maximize the profit? Solve it graphically.

Solution:

Assume number of product I and product II are x and y respectively.

As, profits on every product I and II are 2 and 3 monetary unit. 

As, profits on x number of Product I and y number of Product II are 2x and 3y respectively.

Assume Z is total output daily, 

So,

Z = 2x + 3y

Here, every I and II needs 2 and 4 units of resources A. 

So, x units of product I and y units of product II needs 2x and 4y minutes respectively. 

But, maximum available quantity of resources A is 20 units.

So,

2x + 4y ≤ 20

x + 2y ≤ 10 {First Constraint}

So, each I and II needs 2 and 2 units of resources B. 

So, x units of product I and y units of product II needs 2x and 2y minutes respectively. 

But, maximum available quantity of resources A is 12 units.

So,

2x + 2y ≤ 12

x + y ≤ 6 {Second Constraint}

Here, every unit of product I require 4 units of resources C. 

It is not needed by product II. Thus, x units of product I require 4x units of resource C. 

But, maximum available quantity of resources C is 16 units.

Thus,

4x ≤ 16

x ≤ 4 {Third Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 2x + 3y

Subject to constraints,

x + 2y ≤ 10

x + y ≤ 6

x ≤ 4

x, y ≥ 0  {Since production for I and II can not be less than zero}

Area shown by x + 2y ≤ 10: The line x + 2y = 10 connects the axes at A(10, 0), B(0, 5) respectively.

Area having the origin shows x + 2y ≤ 10 as origin assure x + 2y ≤ 10.

Area shown by x + y ≤ 6: The line x + y = 6 connects the axes at C(6, 0), D(0, 6) respectively. 

Area having the origin shows x + y ≤ 6 as origin assure x + y ≤ 6

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), B(0, 5), G(2, 4), F(4, 2), and E(4, 0).

The values of Z at these corner points are as follows:

Corner PointsZ = 2x + 3y
O0
B15
G16
F14
E8

The maximum value of Z is 16 which is obtained at G (12, 4).

Therefore, the maximum profit is 16 monetary units attained when 2 units of first product and 

4 units of second product were manufactured.

Question 15. A publisher sells a hard cover edition of a text book for Rs 72.00 and paperback edition of the same ext for Rs 40.00. Costs to the publisher are Rs 56.00 and Rs 28.00 per book respectively in addition to weekly costs of Rs 9600.00. Both types require 5 minutes of printing time, although hardcover requires 10 minutes binding time and the paperback requires only 2 minutes. Both the printing and binding operations have 4,800 minutes available each week. How many of each type of book should be produced in order to maximize profit?

Solution:

Assume the sale of hand cover edition be ‘h’ and that of paperback editions be ‘t’.

SP of a hard cover edition of the textbook = Rs 72

SP of a paperback edition of the textbook = Rs 40

Cost to the publisher for hard cover edition = Rs 56

Cost to the publisher for a paperback edition = Rs 28

Weekly cost to the publisher = Rs 9600

Profit to be maximized, Z = (72 – 56)h + (40 – 28)t – 9600

Z = 16h + 12t – 9600

5(h + t) ≤ 4800

10h + 2t ≤ 4800.

The corner points are O(0, 0), B1(0, 960), E1 (360, 600) and F1 (480, 00).

The values of Z at these corner points are as follows:

Corner PointsZ = 16h + 12t – 9600
O-9600
B1920
E3360
F-1920

The maximum value of Z is 3360 which is obtained at E1 (360, 600).

Hence, the maximum profit is 3360 which is attained by selling 360 copies of hardcover edition and 

600 copies paperback edition.

Question 16. A firm manufactures headache pills in two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine; size B contains 1 grain of aspirin, 8 grains of bicarbonate and 66 grains of codeine. It has been found by users that it requires at least 12 grains of aspirin, 7.4 grains of bicarbonate and 24 grains of codeine for providing immediate effects. Determine graphically the least number of pills a patient should have to get immediate relief. Determine also the quantity of codeine consumed by patient.

Solution:

Here the above LPP can be shown as a table below,

 Pill size APill size B 
 XY 
Aspirin2x1y≥ 12
Bicarbonate5x8y≥ 7.4
Codeine1x66y≥ 24
ReliefXYMinimize

Therefore, the formula of the given linear programming problem is 

Max Z = x + y

Subject to constraints,

2x + y  ≥ 12

5x + 8y ≥ 7.4

x + 66y ≥ 24

x, y ≥ 0  {Since production can not be less than zero}

The corner points are B(0, 12), P(5.86, 0.27), E(24, 0)

The values of Z at these corner points are as follows:

Corner PointsZ = x + y
(0, 12)12
(24, 0)24
(5.86, 0.27)6.13

The minimum value of Z is 6.13. 

But,

The area is unbounded thus check whether x + y < 6.13

Thus, we can see that it does not has any common area.

So, x = 5.86, y = 0.27

This is the least quantity of pill A and B.

Codeine quantity = x + 66y = 5.86 + (66 × 0.27) = 24(approx).

Question 17. A chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.

 CompoundMinimum requirement
 AB 
Ingredient C1280
Ingredient D3175
Cost in Rs per kg46 

Solution:

Assume needed quantity of compound A and B are x and y kg.

As, cost of one kg of compound A and B are Rs 4 and Rs 6 per kg. So,

Cost of x kg of compound A and y kg of compound B are Rs 4x and Rs 6y

Respectively.

Assume Z be the total cost of compounds, 

So,

Z = 4x + 6y

As, compound A and B contain 1 and 2 units of ingredient C per kg respectively, 

Thus, x kg of compound A and y kg of compound B contain x and 2y units of 

ingredient C respectively but minimum requirement of ingredient C is 80 units, 

So,

x + 2y ≥ 80 {first constraint}

Hence, compound A and B have 3 and 1 units of ingredient D per kg respectively, 

Hence, x kg of compound A and y kg of compound B have 3x and y units of 

ingredient D respectively but minimum need of ingredient C is 75 units, so,

3x + y ≥ 75 {second constraint}

Therefore, the formula of the given linear programming problem is 

Min Z = 4x + 6y

Subject to constraints,

x + 2y ≥ 80

3x + y ≥ 75

x, y ≥ 0 {Since production can not be less than zero}

Area x + 2y ≥ 80: line x + 2y = 80 connects axes at A(80, 0), B(0, 40) respectively. 

Area not having origin shows x + 2y ≥ 80 as (0, 0) does not assure x + 2y ≥ 80.

Area 3x + y ≥ 75: line 3x + y = 75 connects axes at C(25, 0), D(0, 75) respectively. 

Area not having origin shows 3x + y ≥ 75 as (0, 0) does not assure 3x + y ≥ 75.

Area x,y ≥ 0: it shows first quadrant.

The corner points are D(0, 75), E(14, 33), A(80, 0).

The values at Z at these corner points are as follows:

Corner Point Z = 4x + 6y
D450
E254
A320

The minimum value of Z is 254 which is obtained at E(14, 33).

Therefore, the minimum cost is Rs 254 attained when 14 units of compound A and 

33 units compound B are produced.

Question 18. A company manufactures two types of novelty Souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

Solution:

Assume the company manufacture x souvenirs of Type A and y souvenirs of Type B.

Hence, 

x,  y ≥ 0

Given: The information can be shown in a table as follows:

 Type AType BAvailability
Cutting (min)583 × 60 + 20 = 200
Assembling (min)1084 × 60 = 240

The profit on Type A souvenirs is 50 paisa and on Type B souvenirs is 60 paisa. 

Hence, profit gained on x souvenirs of Type A and y souvenirs of Type B is Rs 0.50x and Rs 0.60y respectively.

Total Profit, 

Z = 0.5x + 0.6y

Therefore, the formula of the given linear programming problem is 

Max Z = 0.5x + 0.6y

Subject to constraints,

5x + 8y ≤ 200

10x + 8y ≤ 240

x ≥ 0, y ≥ 0

Area 5x + 8y ≤ 200: line 5x + 8y = 200 connects axes at A(40, 0), B(0, 25) respectively. 

Area having origin shows the solution of the in-equation 5x + 8y ≤ 200 as (0, 0) assure 5x + 8y ≤ 200.

Area 10x + 8y ≤ 240: line 10x + 8y = 240 connects axes at C(24, 0), D(0, 30) respectively. 

Area having origin shows the solution of the in-equation 10x + 8y ≤ 240 as (0, 0) assures 10x + 8y ≤ 240.

Area x,y ≥ 0: it shows first quadrant.

The corner points of the suitable area are O(0, 0), B(0, 25), E(8, 20), C(24, 0).

The values of Z at these corner points are as follow:

Corner PointsZ = 20x + 30y
O0
B150
E160
C120

The maximum value of Z is obtained at E(8, 20).

Therefore, 8 souvenirs of Type A and 20 souvenirs of Type B must be produced each

day to get the maximum profit of Rs 160.

Question 19. A manufacturer makes two products A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consists of 40 hours of production and weekly turnover must not be less than Rs 10000. If the profit on each of product A is Rs 20 and on product B is Rs 30, then how many of each should be produced so that the profit is maximum. Also, find the maximum profit.

Solution:

Assume x units of product A and y units of product B are manufactured.

The number of units cannot be negative.

So, 

x, y ≥ 0  (always)

As per the question, the given information can be shown as:

 Selling price (Rs)Manufacturing time (hrs)
Product A (x)2000.5
Product B (y)3001

Here, the availability of time is 40 hours and the revenue must be atleast Rs 10000.

Hence further, given, there is a permanent order for 14 units of Product A and 16 units of product B.

Thus, 

The constraints are:

200x + 300y ≥ 10000,

0.5x + y ≤ 40

x ≥ 14

y ≥ 16.

If the profit on every of product A is Rs 20 and on product B is Rs 30. 

Hence, profit gained on x units of product A and y units of product B is Rs 20x and Rs 30y respectively.

Total profit = 20x + 30y which is to be maximized.

Therefore, the formula of the given linear programming problem is 

Max Z = 20x + 30y

Subject to constraints,

200x + 300y ≥ 10000,

0.5x + y ≤ 40

x ≥ 14

y ≥ 16

x, y ≥ 0.

Area 200x + 300y≥  10000: line 200x + 300y = 10000 connects the axes at A(50,0), B(0, 100/3) respectively.

Area not having origin shows 200x + 300y ≥ 10000 as (0, 0) does not assure 200x + 300y ≥ 10000.

Area 0.5x + y ≤ 40: line 0.5x + y = 40 connects the axes at C(80, 0), D(0, 40) respectively.

Area having origin shows 0.5x + y ≤ 40 as (0, 0) assure 0.5x + y ≤ 40.

Area shown by x ≥ 14, x = 14 is the line passes through (14, 0) and is parallel to the Y – axis. 

The area to the right of the line x = 14 will assure the in-equation.

Area shown by y ≥ 16, y = 14 is the line passes through (16, 0) and is parallel to the X – axis. 

The area to the right of the line y = 14 will assure the in-equation.

Area x,y ≥ 0: it shows first quadrant.

The corner points of the suitable area are E(26, 16), F(48, 16), G(14, 33), H(14, 24).

The values of Z at these corner points are as follows:

Corner PointsZ = 20x + 30y
E1000
F1440
G1270
H1000

The maximum value of Z is Rs 1440 which is obtained at F(48,16).

Therefore, the maximum profit is Rs 1440 attained when 48 units of product A and

16 units of product B are manufactured.

Question 20. A manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first types of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of the trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?

Solution:

Assume x trunks of first type and y trunks of second type were manufactured. 

Number of trunks cannot be negative.

Thus, 

x, y ≥ 0 (always)

Given: Information can be shown as

 Machine A (hours)Machine B (hours)
First type (x)33
Second type (y)32
Availability1815

Hence, 

The constraints are,

3x + 3y ≤ 18

3x + 2y ≤ 15.

He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. 

Hence, profit gained by him from x trunks of first type and y trunks of second type is Rs 30x and Rs 25y respectively.

Total profit Z = 30x + 25y that is to be maximized.

Therefore, the formula of the given linear programming problem is 

Max Z = 30x + 25y

Subject to

3x + 3y ≤ 18

3x + 2y ≤ 15

x, y ≥ 0

Area 3x + 3y ≤ 18: line 3x + 3y = 18 connects axes at A(6, 0), B(0, 6) respectively. 

Area having origin shows the solution of the in-equation 3x + 3y ≤ 18 as (0, 0) assures 3x + 3y ≤ 18.

Area 3x + 2y ≤ 15: line 3x + 2y = 15 connects axes at C(5, 0), D(0, 15/2) respectively. 

Area having origin shows the solution of the in-equation 3x + 2y ≤ 15 as (0, 0) assures 3x + 2y ≤ 15.

Area x, y ≥ 0: it shows first quadrant.

The corner points are O(0, 0), B(0, 6), E(3, 3), and C(5, 0).

The values of Z at these corner points are as follows:

Corner PointsZ = 30x + 25y
O0
B150
E165
C150

The maximum value of Z is 165 which is obtained at E(3, 3).

Therefore, the maximum profit is of Rs 165 attained when 3 units of each type of trunk is manufactured.

Question 21. A manufacturer of patent medicines is preparing a production plan on medicines, A and B. There are sufficient raw materials available to make 20000 bottles of A and 40000 bottles of B, but there are only 45000 bottles into which either of the medicines can be put. Further, it takes 3 hours to prepare enough material to fill 1000 bottles of A, it takes 1 hour to prepare enough material to fill 1000 bottles of B and there are 66 hours available for this operation. The profit is Rs 8 per bottle for A and Rs 7 per bottle for B. How should the manufacturer schedule his production in order to maximize his profit?

Solution:

Assume manufacturing of each bottle of A and B are x and y respectively.

As profits on each bottle of A and B are Rs 8 and Rs 7 per bottle respectively. 

Thus, profit on x bottles of A and y bottles of of B are 8x and 7y respectively. 

Assume Z be total profit on bottles 

So,

Z = 8x + 7y

As, it takes 3 hours and 1 hour to prepare enough material to fill 1000 bottles of Type A and Type B respectively, thus, x bottles of A and y bottles of B are preparing is 3x/1000 hours and y/1000 hours respectively, but only 66 hours are available, so, 

3x/1000 + y/1000 ≤ 66

3x + y ≤ 66000

Hence, raw materials accessible to make 2000 bottles of A and 4000 bottles of B but 

there are 45000 bottles in which either of these medicines can be put. 

So,

x ≤ 20000

y ≤ 40000

x + y ≤ 45000

x, y ≥ 0. {Since production of bottles can not be negative}

Therefore, the formula of the given linear programming problem is 

Max Z = 8x + 7y

Subject to constraints,

3x + y ≤ 66000

x ≤ 20000

y ≤ 40000

x + y ≤ 45000

x, y ≥ 0

Area 3x + y ≤ 66000: line 3x + y = 66000 connects the axes at A(22000, 0), B(0, 66000) respectively.

Area containing origin shows 3x + y ≤ 10000 as (0, 0) satisfy 3x + y ≤ 66000

Area x + y ≤ 45000: line x + y = 45000 connects the axes at C(45000, 0), D(0, 45000) respectively.

Area towards the origin will assure the in-equation as (0, 0) assures the in-equation

Area shown by x ≤ 20000,

x = 20000 is the line passes through (20000, 0) and is parallel to the Y – axis. 

The area towards the origin will assure the in-equation.

Area shown by y ≤ 40000,

y = 40000 is the line passes through (0,40000) and is parallel to the X – axis. 

The area towards the origin will assure the in-equation.

Area x,y ≥ 0: it shows first quadrant.

The corner points are O(0, 0), B(0, 40000), G(10500, 34500), H(20000, 6000), A(20000, 0).

The values of Z at these corner points are,

Corner PointsZ = 8x + 7y
O0
B280000
G325500
H188000
A160000

The maximum value of Z is 325500 which is obtained at G(10500, 34500).

Therefore, the maximum profit is Rs 325500 attained 

when 10500 bottles of A and 34500 bottles of B are manufactured.

Question 22. An aeroplane can carry a maximum of 200 passengers. A profit of Rs 400 is made on each first class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats of first class. However, at least 4 times as many passengers prefer to travel by economy class to the first class. Determine how many each type of tickets must be sold in order to maximize the profit for the airline. What is the maximum profit?

Solution:

Assume needed number of first class and economy class tickets be x and y respectively.

Each ticket of first class and economy class made profit of Rs 400 and Rs 600 respectively.

Thus, x ticket of first class and y tickets of economy class make profit of Rs 400x and Rs 600y respectively.

Assume the total profit be Z and it is shown by

Z = 400x + 600y

Given, aeroplane can carry a minimum of 200 passengers, 

So,

x + y ≤ 200

Given airline reserves at least 20 seats for first class, 

So,

x ≥ 20

Also, at the minimum 4 times as many passengers choose to travel by economy class to the first class, 

So

y ≥ 4x

Therefore, the formula of the given linear programming problem is 

Max Z = 400x + 600y

Subject to constraints

x + y ≤ 200

y ≥ 4x

x ≥ 20

x, y ≥ 0 {since seats in both the classes can not be zero}

Area shown by x + y ≤ 200: the line x + y = 200 connects the axes at A(200, 0), B(0, 200). 

Area having origin shows x + y ≤ 200 as (0, 0) assures x + y ≤ 200.

Area shown by x ≥ 20: line x = 20 passes through (20, 0) and is parallel to y-axis. 

The area to the right of the line x = 20 will assure the in-equation x ≥ 20

Area shown by y ≥ 4x: line y = 4x passes through (0, 0). 

The area over the line y = 4x will assure the in-equation y ≥ 4x

Area x, y ≥ 0: it shows the first quadrant.

The corner points are C(20, 80), D(40, 160), E(20, 180).

The values of Z at these corner points are as follows:

Corner Points Z = 400x + 600y
O0
C56000
D112000
E116000

The maximum value of Z is attained at E(20, 180).

Therefore, the maximum profit is Rs 116000 attained when 20 first class tickets and 180 economy class tickets are sold.

Question 23. A gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6% phosphoric acid and type II fertilizer, which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg, determine how many kilograms of each fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Solution:

Assume x kg of Type I fertilizer and y kg of Type II fertilizers are supplied.

The quantity of fertilizers can not be negative.

So, 

x, y ≥ 0 (always)

A gardener has a supply of fertilizer of type I which contains of 10% nitrogen and 

Type II contains of 5% nitrogen, and he requires at least 14 kg of nitrogen for his crop.

Thus,

(10 × 100) + (5 × 100) ≥ 14

Or, 

10x + 5y ≥ 1400

A gardener has a supply of fertilizer of type I which contains of 6% phosphoric acid and 

Type II contains of 10% phosphoric acid, and he requires at least 14 kg of phosphoric acid for his crop.

Thus,

(6 × 100) + (10 × 100) ≥ 14

Or, 

6x + 10y ≥ 1400

Hence,

Subject to constraints are,

10x + 5y ≥ 1400

6x + 10y ≥ 1400

Assuming that the Type I fertilizer costs 60 paise per kg and Type II fertilizer costs 40 paise per kg. 

Thus, the cost of x kg of Type I fertilizer and y kg of Type II fertilizer is Rs0.60x and Rs 0.40y respectively.

Assume the Total cost is Z

So,

Z = 0.6x + 0.4y is to be minimized.

Therefore, the formula of the given linear programming problem is 

Min Z = 0.6x + 0.4y

Subject to the constraints,

10x + 5y ≥ 1400

6x + 10y ≥ 1400

x, y ≥ 0

The area shown by 6x + 10y ≥ 1400: line 6x + 10y = 1400 passes through A(700/3, 0) and B(0, 140). 

The area which doesn’t consist the origin shows the solution of the in-equation 6x + 10y ≥ 1400

Since, (0,0) doesn’t assure the in-equation 6x + 10y ≥ 1400

Area shown by 10x + 5y ≥ 1400: line 10x + 5y = 1400 passes through C(140, 0) and D(0, 280). 

The area that doesn’t consist the origin shows the solution of the in-equation 10x + 5y ≥ 1400

As (0,0) doesn’t assure the in-equation 10x + 5y ≥ 1400

The area, x,y ≥ 0: shows the first quadrant.

The corner points are D(0, 280), E(100, 80), A(700/3, 0)

The values of Z at these points are as follows:

Corner Points Z = 0.6x + 0.4y
O0
D112
E92
F140

The minimum value of Z is Rs 92 which is obtained at E(100, 80)

Thus, the minimum cost is Rs92 attained when 100 kg of Type I fertilizer and 80 kg of Type II fertilizer is supplied.

Question 24. Anil wants to invest at most Rs 12000 in Saving Certificates and National Saving Bonds. According to rules, he has to invest at least Rs 2000 in Saving Certificates and at least Rs 4000 in National Saving Bonds. If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum, how much money should he invest to earn maximum yearly income? Find also his maximum yearly income.

Solution:

Let Anil invests Rs x and Rs y in saving certificate (SC) and National saving bond (NSB) respectively.

Since, the rate of interest on SC is 8% annual and on NSB is 10% annual. 

So, interest on Rs x of SC is 8x/100 and Rs y of NSB is 10y/100 per annum.

Assure Z be total interest earned so,

Z = 8x/100 + 10y/100

Given: He wants to invest Rs 12000 is total

x + y ≤ 12000

According to the rules he has to invest at least Rs 2000 in SC and at least Rs 4000 in NSB.

x ≥ 2000

y ≥ 4000

Therefore, the formula of the given linear programming problem is 

Max Z = 8x/100+ 10y/100  

Subject to constraints

x ≥ 2000

y ≥ 4000

x + y ≤ 12000

x,y ≥ 0

The area shown by x ≥ 2000: line x = 2000 is parallel to the y – axis and passes through (2000, 0).

The area not having the origin shows x ≥ 2000

As (0, 0) doesn’t assure the in-equation x ≥ 2000

The area shown by y ≥ 4000: line y = 4000 is parallel to the x – axis and passes through (0, 4000).

The region not containing the origin represents y ≥ 4000

As (0, 0) doesn’t assure the in-equation y ≥ 4000

Area shown by x + y ≤ 12000: line x + y = 12000 connects axes at A(12000, 0) and B(0, 12000) respectively. 

The area which have the origin shows the solution set of x + y ≤ 12000

As (0, 0) assure the in-equality x + y ≤ 12000.

Area x, y ≥ 0 is shown by the first quadrant.

The corner points are E(2000, 10000), C(2000, 4000), D(8000, 4000).

The values of Z at these corner points are as follows:

Corner PointsZ = 8x/100+ 10y/100  
O0
E1160
D1040
C560

The maximum value of Z is Rs 1160 which is obtained at E(2000,10000).

Thus, the maximum earning is Rs1160 attained when Rs 2000 were invested in SC and Rs 10000 in NSB.

Question 25. A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs 25 per tree. When fully grown, type A produces an average of 20 kg of fruit which can be sold at a profit of Rs 2.00 per kg and type B produces an average of 40 kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?

Solution:

Assume the needed number of trees of Type A and B be Rs x and Rs y respectively.

Number of trees cannot be negative.

x, y ≥ 0. (always)

To plant tree of Type A requires 10 sq. m and Type B needs 20 sq. m of ground per tree. 

And it is given that a man owns a field of area 1000 sq. m. 

Therefore,

10x + 20y ≤ 1000

x + 2y ≤ 100

Type A costs Rs 20 per tree and Type B costs Rs 25 per tree. 

Therefore, x trees of type A and y trees of type B cost Rs 20x and Rs 25y respectively. 

A man has a sum of Rs 1400 to purchase young trees.

20x + 25y ≤ 1400

4x + 5y ≤ 280

Therefore, the mathematical formulation of the given LPP is

Max Z = 40x – 20x + 60y – 25y = 20x + 35y

Subject to constraints;

x + 2y ≤ 100

4x + 5y ≤ 280

x, y ≥ 0

Area 4x + 5y ≤ 280: line 4x + 5y ≤ 280 connects axes at A1(70, 0), B1(0, 56) respectively.

The area having origin shows 4x + 5y ≤ 280 as (0, 0) assure 4x + 5y ≤ 280.

Area x + 2y ≤ 100: line x + 2y = 100 connects axes at A2(100, 0), B2(0, 50) respectively.

Area having origin shows x + 2y ≤ 100 as (0, 0) assure x + 2y ≤ 100

Area x, y ≥ 0: it shows the first quadrant.

The corner points are A1(70, 0), P(20, 40), B2(0, 50)

The values of Z at these corner points are as follows:

Corner PointsZ = 20x + 35y
O0
A11750
P1800
B21400

The maximum value of Z is 1800 which is obtained at P(20, 40).

Thus, the maximum profit is Rs 1800 attained when Rs 20 were 

involved in Type A and Rs 40 were involved in Type II.

Question 26. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and sprayer. It takes 2 hours on the grinding/cutting machine and 3  hours on the sprayer to manufacture a pedestal lamp while it takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at most 20 hours and the grinding/cutting machine for atmost 12 hours. The profit from the sale of a lamp is 5.00 and a shade is 3.00. Assuming that the manufacturer sell all the lamps and shades that he produces, howshould he schedule his daily production in order to maximise his profit?

Solution:

Assume x units of pedestal lamps and y units of wooden shades are manufactured on a day to maximise the profit.

As a pedestal lamp needs 2 hours on the grinding/cutting machine and a wooden shade needs 1 hour on the grinding/cutting machine, 

Hence, the total hours needs for grinding/cutting x units of pedestal lamps and 

y units of wooden shades are (2x + y). 

Thus, 

The grinding/cutting machine is accessible for at maximum 12 hours on a day.

Therefore,

2x + y ≤ 12 

As well as,

A pedestal lamp needs 3 hours on the sprayer and a wooden shade needs 2 hours on the sprayer.  

Therefore, the total hours needed for spraying x units of pedestal lamps and

y units of wooden shades are (3x + 2y). 

As, the sprayer is accessible for at most 20 hours on a day.

Therefore,

3x + 2y ≤ 20

The profit from the sale of a pedestal lamp is 5.00 and a wooden shade is 3.00. 

Thus, the total profit from the sale of x units of pedestal lamps and y units of wooden shades is (5x + 3y). 

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 5x + 3y

Subject to the constraints 

2x + x ≤ 12

3x + 2y ≤ 20

 x, y ≥ 0

The suitable area determined by the given constraints can be diagrammatically shown as,

The coordinates of the corner points of the feasible region are O(0, 0), A(6, 0), B(4, 4) and C(0, 10). 

The value of the objective function at these points are given in the following table.

Corner Point Z = 5x + 3y
0, 05 × 0 + 3 × 0 = 0
6, 05 × 6 + 3 × 0 = 30
4, 45 × 4 + 3 × 4 = 32   →Maximum
0, 105 × 0 + 3 × 10 = 30

The maximum value of Z is 32 at x = 4, y = 4.

Therefore, the manufacturer must manufacture 4 pedestal lamps and 4 wooden shades to maximise his profit. 

The maximum profit of the manufacturer is 32 on a day.

Question 27. A producer has 30 and 17 units of labour and capital respectively which he can use to  produce two type of goods x and y. To produce one unit of x, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital is required to produce one unit of y. If x and y  are priced at Rs 100 and Rs 120 per unit respectively, how should be producer use his resources to maximize the total revenue? Solve the problem graphically.

Solution:

Let x1 and y1 units of goods x and y were produced respectively.

Number of units of goods cannot be negative.

Therefore, x1, y1 ≥ 0

To manufacture one unit of x, 2 units of labour and for one unit of y, 3 units of labour are needed.

2x1 + 3y1 ≤ 30

To manufacture one unit of x, 3 units of capital is needed and 

1 unit of capital is needed to manufacture one unit of y.

3x1 + y1 ≤ 17

If x and y are priced at Rs 100 and Rs 120 per unit respectively, 

Thus, price of x1 and y1 units of goods x and y is Rs 100x1 and Rs 120y1 respectively.

Total revenue

Z = 100x1 + 120y1 which is to be maximised.

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 100x1 + 120y1

Subject to 

2x1 + 3y1 ≤ 303x1 + y1 ≤ 17

x, y ≥ 0

In the starting we will convert in-equations into equations as follows:

2x1 + 3y1 = 30, 

3x1 + y1 = 17, 

x = 0 and 

y = 0

Area shown by 2x1 + 3y1 ≤ 30:

The line 2x1 + 3y1 = 30 meets the coordinate axes at A(15, 0) and B(0, 10) respectively. 

After connecting these points we will get the line 2x1 + 3y1 = 30. 

Thus, (0, 0) assures the 2x1 + 3y1 = 30. 

Hence, the area that have the origin shows the solution set of the in-equation 2x1 + 3y1 ≤ 30.

Area shown by 3x1 + y1 ≤ 17:

The line 3x1 + y1 = 17 connects the coordinate axes at C ( 17/3 , 0) and D(0, 17) respectively. 

After connecting these points we will get the line

3x1 + y1 = 17. 

Here, (0, 0) assures the in-equation 3x1 + y1 ≤ 17. 

Thus, the area that have the origin shows the solution set of the in-equation 3x1 + y1 ≤ 17.

Area shown by x1 ≥ 0 and y1 ≥ 0:

As, all the points in the first quadrant assures these in-equations. 

Thus, the first quadrant is area shown by the in-equations x ≥ 0, and y ≥ 0.

The suitable area is determined by the system of constraints 

2x1 + 3y1 ≤ 30, 

3x1 + y1 ≤ 17, 

x ≥ 0 and 

y ≥ 0 are as follows

The corner points are B(0, 10), E(3, 8) and C( 17/3 , 0 ) . The values of Z at these corner points are as follows

Corner pointZ= 100x1 + 120y1
B1200
E1260
C1700/3

The maximum value of Z is 1260 that is obtained at E(3, 8).

Therefore, the maximum revenue is Rs 1260 attained when 3 units of x and 8 units of y were manufactured.

Question 28. A firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of type A and Rs 3 per unit of type B. Each product is processed on two machines M1 and M2. One unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2,  whereas one unit of type B requires one minute of processing time on M1 and one minute on M2 Machines M1 and Mare respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce a day in order to maximize the profit. Solve the problem graphically.

Solution:

Assume x units of product A and y units of product B were produced. 

Number of products cannot be negative.

Thus, 

x, y ≥ 0 (always) 

As in the question, the given information can be shown as

 Time on M1 (minutes)Time on M2 (minutes)
Product A(x)12
Product B(y)11
Availability300360

The constraints are

x + y ≤ 300, 

2x + y ≤ 360

Firm produces two types of products A and B and sells them at a profit of Rs 5 per unit oftype A and 

Rs 3 per unit of type B.

Thus, x units of product A and y units of product B costs Rs 5x and Rs3y respectively. 

Total profit is Z = 5x + 3y which has to be maximised

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 5x + 3y

Subject to

x + y ≤ 300, 

2x + y ≤ 360 

x, y ≥ 0

In the starting we will convert in-equations into equations as follows:

x + y = 300,  

2x + y = 360, 

x = 0 and 

y = 0

The area shown by x + y ≤ 300:

The line x + y = 300 connects the coordinate axes at A1(300, 0) and B1(10, 300) respectively. 

After connecting these points we will get the line x + y = 30. 

As (0, 0) assures the x + y = 30. 

Hence, the area that have the origin shows he solution set of the in-equation

x + y ≤ 300.

The area shown by 2x + y ≤ 360:

The line 2x + y = 360 connects the coordinate axes at C1(180, 0) and D1(10, 360) respectively. 

After connecting these points we will get the line

2x + y = 360. 

As (0, 0) assures the in-equation 2x + y ≤ 360. 

Thus, the area which have the origin shows the solution set of the in-equation 2x + y ≤ 360.

The area shown by x ≥ 0 and y ≥ 0:

Thus, all the points in the first quadrant assures these in-equations. 

Thus, the first quadrant is the area shown by the in-equations x ≥ 0, and y ≥ 0.

The suitable area shown by the system of constraints x + y ≤ 300, 

2x + y ≤ 360, 

x ≥ 0 and 

y ≥ 0 are as follows;

The corner points are O(0, 0), B1(0, 300), E1(60, 240) and C1(180, 0). The values of Z at these corner points are as follows

Corner pointZ= 5x + 3y
O0
B1900
E11020
C1900

The maximum value of Z is Rs 1020 that is obtained at B1(60, 240).

Therefore, the maximum profit is Rs 1020 attained when 60 units of product A and 

240 units of product B were produced.

Question 29. A small firm manufactures items A and B. The total number of items A and B that it can manufacture in a  day is at the most 24. Item A takes one hour to make while item B  takes only half an hour. The maximum time available per day is 16 hours.  If the profit on one unit of item A be  Rs 300 and one unit of item B be Rs 160, how many of each type of item be produced to maximize the profit? Solve the problem graphically.

Solution:

Assume the firm manufacture x items of A and y items of B per day. 

Number of items cannot be negative.

Thus, 

x, y ≥0 (always)

It is given that the total number of items manufactured per day is at most 24.

Therefore, 

x + y ≤ 24

Item A requires 1 hour to make and item B requires 0.5 hour to make. 

The maximum number of hours available per day is 16 hours.

Therefore, 

x + 0.5y ≤ 16

If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160.

Thus, profit gained on x items of A and y items of B is Rs 300x and Rs 160y respectively.

Therefore, 

Total profit is Z = 300x + 160y

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 300x + 160y 

Subject to the constraints 

x+ y ≤ 24

x + 0.5y ≤ 16

x, y ≥0

At starting, we will convert in-equations into equations as follows:

x + y = 24, 

x + 0.5y = 16, 

x = 0 and 

y = 0

The area shown by x + y ≤ 24:

The line x + y = 24 connects the coordinate axes at A1(24, 0) and B1(0, 24) respectively. 

By connecting these points we will get the line x + y = 24.

As, (0, 0) assures the x + y = 24. 

Thus, the area having the origin shows the solution set of the in-equation

x + y ≤ 24.

The area shown by x + 0.5y ≤ 16:

The line x + 0.5y = 16 connects the coordinate axes at C1(16, 0) and D1(0, 32) respectively. 

After connecting these points we will get the line 

x + 0.5y = 16. 

As (0, 0) assure the in-equation x + 0.5y ≤ 16. 

Thus,the area that have the origin shows the solution set of the in-equation 

x + 0.5y ≤ 16.

The area shown by x ≥ 0 and y ≥ 0:

As, all the points in the first quadrant assures these in-equations. 

Thus, the first quadrant is the area shown by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints 

x + y ≤ 24, 

x + 0.5y ≤ 16, 

x ≥ 0 and 

y ≥ 0 are as follows

The corner points are O(0, 0) ,C1(16, 0), E1(8, 16) and B1(0, 24). 

The value of Z at these corner points are as follows:

Corner PointZ = 300x + 160y
O(0, 0)0
C(16, 0)4800
E1(8, 16)4960
B1(0, 24)3840

Therefore, the maximum value of Z is 4960 at E1(8, 16).

Therefore, 8 units of item A and 16 units of item B should be manufactured per day to maximise the profits.

Question 30. A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?

Solution:

Assume x toys of type A and y toys of type B were manufactured. 

The given information can be shown in the tabulated as follows:

 Cutting time (minutes)Assembling time (minutes)
Toy A(x)510
Toy B(y)88
Availability180240

The constraints are

5x + 8y ≤ 180

10x + 8y ≤ 240

The profit is Rs 50 each on type A and Rs 60 each on type B. 

Thus, profit gained on x toys of type A and y toys of type B is Rs 50x and Rs 60 y respectively.

Total profit is Z = 50x + 60y

Therefore, the mathematical formulation of the given LPP is 

Maximise Z = 50x + 60y

Subject to

5x + 8y ≤ 180

10x + 8y ≤ 240

At first we will convert in-equations into equations as follows: 

5x + 8y = 180, 

10x + 8y = 240, 

x = 0 and 

y = 0

The area shown by 5x + 8y ≤ 180:

The line 5x + 8y = 180 connects the coordinate axes at A1(36, 0) and B1 (0, 45/2) respectively. 

After connecting these points we will get the line 5x +8y = 180. 

As, (0, 0) assures the 5x + 8y = 180. 

Thus, the area that have the origin shows the solution set of the in-equation 5x + 8y ≤ 180.

The area shown by 10x + 8y ≤ 240:

The line 10x + 8y = 240 meets the coordinate axes at C1(24, 0) and D1(0, 30) respectively.

After connecting these points we will get the line 10x + 8y = 240. 

As (0, 0) assures the in-equation 10x + 8y ≤ 240.

Thus, the area that have the origin shows the solution set of the in-equation 10x + 8y ≤ 240.

The area shown by x ≥ 0 and y ≥ 0:

As, all the points in the first quadrant assures these in-equations. 

Thus, the first quadrant is the area shown by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints 

5x + 8y ≤ 180, 

10x + 8y ≤ 240, 

x ≥ 0 and 

y ≥ 0 are as follows. 

The suitable area is shown in the figure

The corner points are  B1(0, 45/2), E1(12, 15) and C1(24, 0).

The values of Z at the corner points are

Corner pointsZ = 50x + 60y
O0
B11350
E11500
C11200

The maximum value of Z is 1500 which is at E1(12, 15).

Therefore, for maximum profit, 12 units of toy A and 15 units of toy B should be manufactured.

Question 31.  A company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments.  The maximum capacity of the first department is 60 hours a week and that of the other department is 48 hours per week. The product of each unit of article  A needs 4 hours in assembly and 2 hours in finishing and that of each unit of B  needs 2 hours in assembly and 4 hours in finishing. If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B,  find the number of units of A and B to be produced per week in order to have maximum profit.

Solution:

Assume x units and y units of articles A and B are produced respectively. 

Number of articles cannot be negative.

Therefore, 

x, y ≥ 0 (always)

The product of every unit of article A needs 4 hours in assembly and that of article B needs 2 hours in assembly and the maximum capacity of the assembly department is 60 hours a week.

4x + 2y ≤ 60

The product of each unit of article A needs 2 hours in finishing and that of article B needs 

4 hours in assembly and the maximum capacity of  the finishing department is 48 hours a week.

2x + 4y ≤ 48

If the profit is Rs 6 for every unit of A and Rs 8 for every unit of B. 

Therefore, profit gained from x units and y units of articles A and 

B respectively is Rs 6x and Rs 8y respectively.

Total revenue  Z = 6x + 8y that is to be maximised.

Therefore, the mathematical formulation of the given LPP is 

Max Z = 6x + 8y

Subject to constraints

2x + 4y ≤ 48

4x + 2y ≤ 60

x, y ≥ 0

First we will convert in-equations into equations so follows:

2x + 4y = 48, 

4x + 2y = 60, 

x = 0 and y = 0 

Area shown by 2x + 4y ≤ 48:

The line 2x + 4y = 48 connects the coordinate axes at A(24, 0) and B(0, 12) respectively. 

By connecting these points we will get the line 2x + 4y = 48. 

So, (0, 0) assure the 2x + 4y = 48. 

So, the area which haves the origin shows the solution set of the in-equation 2x + 4y ≤ 48. 

Area shows by 4x + 2y ≤ 60:

The line 4x + 2y = 60 connects the coordinate axes at C(15, 0) and D(0, 30) respectively. 

By connecting these points we will get the line 4x + 2y = 60. 

So, (0, 0) assure the in-equation 4x + 2y ≤ 60. 

So, the area which haves the origin shows the solution set of the in-equation 4x + 2y ≤ 60.

Area shows by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations. 

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints 

2x + 4y ≤ 48, 

4x + 2y ≤ 60, 

x ≥ 0 and 

y ≥ 0 are so follows

The corner points are O(0, 0), B(0, 12), E(12, 6) and C(15, 0). 

The values of Z at these corner points are so follows

Corner PointsZ = 6x + 8y
O0
B96
E120
C90

The maximum value of Z is 120 which is obtained at E(12, 6).

Therefore, the maximum profit is Rs 120 will get when 12 units of article A and 

6 units of article B were manufactured

Question 32. A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? solve the problem graphically.

Solution:

Assume x and y be the number of items of A and B that should be produced each day to maximize the profit. 

Number of items cannot be negative.

Therefore,

 x, y ≥ 0

It is also given that the firm can produce at most 24 items in a day.

Therefore,

x+ y ≤ 24

Also, the time needed to make an item of A is one hour and time needed to make an item of B is half an hour.

Therefore, the time needed to produce x items of A and y items of B is hours. 

However, the maximum time available in a day is 16 hours.

Therefore,

x + y/2 ≤ 16

It is given that the profit on an item of A is Rs 300 and on one item of B is Rs 160. 

Therefore, the profit gained from x items of A and y items of B is Rs 300x and Rs 160y respectively. 

Total profit Z = 300x + 160y

Therefore, the mathematical formulation of the given LPP is 

Maximize Z = 300x + 160y 

Subject to constraints:

x + y ≤ 24,

x+\frac{1}{2}y\le16

x ≥ 0, 

y ≥ 0

First we will convert in-equations into equations so follows:

x + y = 24, 

x + y/2 ≤ 16

x = 0 and y = 0 

Area shows by x + y ≤ 24:

The line x + y = 24 connects the coordinate axes at A(24, 0) and B(0, 24) respectively. 

By connecting these points we will get the line x + y = 24. 

So, (0, 0) assure the x + y = 24. 

So, the area which have the origin shows the solution set of the in-equation x + y ≤ 24.

Area shows by x + y/2 ≤ 16

The line x + y/2 ≤ 16 connects the coordinate axes at C(16, 0) and D(0, 32) respectively. 

By connecting these points we will get the line x + y/2 = 16

So, (0, 0) assure the in-equation x + y/2 ≤ 16. 

So, the area which haves the origin shows the solution set of the in-equation x + y/2 ≤ 16.

Area shows by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations. 

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints x + y ≤ 24, x + y/2 ≤ 16, x ≥ 0 and 

y ≥ 0 are so follows. 

The suitable area is shown in the figure.

In the above graph, the shaded area is the suitable area. The corner points are O(0, 0), C(16, 0), E(8, 16), and B(0, 24).

The values of the objective function Z at corner points of the suitable area are given in the following table:

Corner Points Z = 300x + 160y
 O(0, 0) 0
C(16, 0)4800
E(8, 16)4960  Maximum
 B(0, 24)3840

So, Z is maximum at x = 8 and y = 16 and the maximum value of Z at this point is 4960.

Therefore, 8 items of A and 16 items of B should be produced in order to maximize the profit and 

the maximum profit is Rs 4960.

Question 33. A company sells two different products, A and B. The two products are produced in a  common production process, which has a total capacity of  500 man-hours.  It takes 5 hours to produce a unit of A  and 3 hours to produce a  unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for  B is 125.  If the profit is Rs 20  per unit for the product A and Rs 15 per unit for the product B, how many units of each product should be sold to maximize profit?

Solution:

Assume x units of product A and y units of product B were manufactured. 

So, x ≥ 0, y ≥ 0

It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B.

The two products are produced in a common production process, which has a total capacity of 500 man-hours

5x + 3y ≤ 500

The maximum number of unit of A that can be sold is 70 and that for B is 125.

 x ≤ 70y ≤ 125

If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B

Therefore, profit x units of product A and y units of product B is Rs 20x and Rs 15yrespectively.

Total profit = Z = 20x + 15y

Therefore, the mathematical formulation of the given LPP is 

Max Z = 20x + 15y

Subject to constraints

5x + 3y ≤ 500,

x ≤ 70,

y ≤ 125,

x ≥ 0, y ≥ 0

First we will convert in-equations into equations so follows: 

5x + 3y = 500, 

x = 70, 

y = 125, 

x = 0 and y = 0

Area shows by 5x + 3y ≤ 500

The line 5x + 3y = 500 connects the coordinate axes at A(100, 0) and B(0, 500/3) respectively. 

By connecting these points we will get the line 5x +3y = 500. 

So, (0, 0) assure the 5x + 3y = 500. 

So, the area which haves the origin shows the solution set of the in-equation 5x + 3y ≤ 500.

Area shows by x ≤ 70:

The line x = 70 is the line passes through C(70, 0) and is parallel to Y axis. 

The area to the left of the line x = 70 will assure the in-equation x ≤ 70.

Area shows by y ≤ 125:

The line y = 125 is the line passes through D(0, 125) and is parallel to X axis. 

The area below the line y = 125 will assure the in-equation y ≤ 125.

Area shows by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations. 

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints 

5x + 3y ≤ 500, 

x ≤ 70, 

y ≤ 125, 

x ≥ 0 and y ≥ 0 are so follows

The corner points are O(0, 0), D(0, 125), E(25, 125), F(70, 50) and C(70, 0). 

The values of Z at the corner points are

Corner pointsZ = 20x + 15y
O0
D1875
E2375
F2150
C1400

The maximum value of Z is 2375 which is at E(25, 125).

Therefore, the maximum profit is Rs 2375, 25 units of A and 125 units of B should be manufactured

Question 34. A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes need 4 sq. meter per box while the small boxes need 3 sq. meter per box. The manufacturer is needed to make at least three large boxes  and at least twice so many small boxes so large boxes.  If  60  sq. meter of cardboard is in stock,  and if the profits on the large and small boxes are  Rs 3 and  Rs 2 per box,  how many of every should be made in order to maximize the total profit?

Solution:

Assume x large boxes and y small boxes be manufactured. Number of boxes cannot be negative.

Therefore, 

x ≥ 0, y ≥ 0 (always)

The large boxes need 4 sq. meter per box while the small boxes need 3 sq. meter per box and 

if 60 sq. meter of cardboard is in stock. 4x + 3y ≤ 60

The manufacturer is needed to make at least three large boxes and at least twice

so many small boxes so large boxes.

x ≥ 3y ≥ 2x

If the profits on the large and small boxes are Rs 3 and Rs 2 per box. 

Thus, profit gained by him on x large boxes and y small boxes is Rs 3x and Rs 2y respectively.

Total profit = Z = 3x + 2y

Therefore, the mathematical formulation of the given LPP is 

Max Z = 3x + 2y

Subject to constraints

4x + 3y ≤ 60

x ≥ 3

y ≥ 2x 

x ≥ 0, y ≥ 0

First we will convert in-equations into equations so follows:

4x + 3y = 60, 

x = 3, 

y = 2x, 

x = 0 and 

y = 0

Area shown by 4x + 3y ≤ 60:

The line 4x + 3y = 60 connects the coordinate axes at A(15, 0) and B(0, 20) respectively. 

By connecting these points we will get the line 4x + 3y = 60. 

So, (0, 0) assure the 4x + 3y = 60. 

So, the area which haves the origin shows the solution set of the in-equation 4x + 3y ≤ 60.

Area shown by x ≥ 3:

The line x = 3 is the line passes through 3, 0 and is parallel to Y axis. 

The area to the right of the line x = 3 will assure the in-equation x ≥ 3.

Area shown by y ≥ 2x:

The line y = 2x is the line that passes through (0, 0). 

The area above the line y = 2x will assure the in-equation y ≥ 2x. 

Like if we take an example taking a point 5, 1 below the line y = 2x .

Here, 1 < 10 which does not assure the in-equation y ≥ 2x. 

Therefore, the area above the line y = 2x will assure the in-equation y ≥ 2x.

Area shown by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations. 

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints 

4x + 3y ≤ 60, 

x ≥ 3, 

y ≥ 2x, 

x ≥ 0 and y ≥ 0 are as follows

The corner points are E(3, 16), D(6, 12) and C(3, 6). 

The values of Z at the corner points are

Corner pointsZ = 3x + 2y
E41
D42
C21

The maximum value of Z is 42 which is at D(6, 12).

Therefore, for maximum profit is Rs 42, 6 units of large boxes and

12 units of smaller boxes must be manufactured.

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