RD Sharma Class 12 Ex 30.5 Solutions Chapter 30 Linear Programming

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter30
Exercise30.5
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 30.5 Solutions Chapter 30 Linear Programming

Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.5

Question 1. Two godowns A and B have gain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

Transportation cost per quintal (in Rs)
From/ ToAB
D64
E32
F2.503

How should the supplies be transported in order that the transportation cost is minimum? what is the minimum cost?

Solution:

Let us assume godown A supply x and y quintals of grain to the shops D and E.

Afterwards, (100 – x – y) will be supplied to shop F.

The requirement at shop D is 60 quintals 

Since, x quintals are transported from godown A.

Therefore, the remaining (60 − x) quintals will be transported from godown B.

Similarly, 

(50 − y) quintals and 40 − (100 − x − y) i.e. (x + y − 60) quintals will be transported from godown B to shop E and F respectively.

The diagrammatic representation of the given problem:

x ≥ 0 , y ≥ 0 and 100 – x – y ≥ 0  

⇒ x ≥ 0 , y ≥ 0 , and x + y ≤ 100

60 – x ≥ 0 , 50 – y ≥ 0 , and x + y – 60 ≥ 0

⇒ x ≤ 60 , y ≤ 50 , and x + y ≥ 60  

Total transportation cost Z is given by,

Z = 6x + 3y + 2.5(100 – x – y) + 4(60 – x) + 2(50 – y) + 3( x + y – 60)

= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180

= 2.5x + 1.5y + 410

Thus, the required mathematical formulation of linear programming is:

Minimize Z = 2.5x + 1.5y + 410  

subject to the constraints,

x + y ≤ 100

x ≤ 60

y ≤ 50

x + y ≥ 60

x, y ≥ 0

The feasible region obtained by the system of constraints is:

The corner points are A(60, 0), B(60, 40), C(50, 50), and D(10, 50).

The values of Z at these corner points are given below.

Corner pointZ = 2.5x + 1.5y + 410
A (60, 0)560
B (60, 40)620
C (50, 50)610
D (10, 50)510 -> minimum

Therefore, 

The minimum value of Z is 510 at D (10, 50).

Hence, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively.

From B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.

The minimum cost is Rs 510.

Question 2. A medical company has factories at two places, A and B. From these places, supply is made to each of its three agencies situated at P, Q and R. The monthly requirements of the agencies are respectively 40, 40 and 50 packets of the medicines, while the production capacity of the factories, A and B are 60 and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given below:

Transportation cost per packet (in Rs)
To\FromAB
P54
Q42
R35

How many packets from each factory be transported to each agency so that the cost of transportation is minimum? Also, find the minimum cost?

Solution:

The diagrammatic representation of the given problem:

Assume x and y packets be transported from factory A to the agencies P and Q respectively.  

After that, 

[60 − (x + y)] packets be transported to the agency R.

First constraint ⇢  x, y ≥ 0 and

Second constraint ⇢ 60 − (x + y) ≥ 0

(x + y) ≤ 60

The requirement at agency P is 40 packets. 

Since, x packets are transported from factory A,  

Therefore, the remaining (40 − x) packets are transported from factory B.

Similarly, 

(40 − y) packets are transported by B to Q and 50− [60 − (x + y)] 

i.e. (x + y − 10) packets will be transported from factory B to agency R respectively.

Number of packets cannot be negative.

Hence,

Third constraint ⇢ 40 – x ≥ 0

=> x ≤ 40

Fourth constraint ⇢ 40 – y ≥ 0

=> y ≤ 40

Fifth constraint ⇢ x + y – 10 ≥ 0

=> x + y ≥ 10

Thus, 

Costs of transportation of each packet from factory A to agency P, Q, R are Rs 5, 4, 3.

Costs of transportation of each packet from factory B to agency P, Q, R are Rs 4, 2, 5.

Let total cost of transportation be Z.

Z = 5x + 4y + 3[60 – x + y] + 4(40 – x) + 2(40 – y) + 5(x + y – 10]

= 3x + 4y + 10

Hence, 

The required mathematical formulation of linear programming is:

Minimize Z = 3x + 4y + 370

subject to constraints,

x + y ≤ 60

x ≤ 40

y ≤ 40

x + y ≥ 10

where, x, y ≥ 0

Let us convert inequations into equations as follows:

x + y = 60, x = 40, y = 40, x + y = 10, x = 0 and y = 0

Region represented by x + y ≤ 60:

The line x + y = 60 meets the coordinate axes at A1 (60, 0) and B1 (0, 60) respectively. 

Region containing origin represents x + y ≤ 60 as (0, 0) satisfies x + y ≤ 60.

Region represented by x ≤ 40:

The line x = 40 is parallel to y-axis, meets x-axis at A2 (40, 0). Region containing origin represents x ≤ 40 as (0, 0) satisfies x ≤ 40.

Region represented by y ≤ 40:

The line y = 40 is parallel to x-axis, meets y-axis at B2 (0, 40). 

Region containing origin represents y ≤ 40 as (0, 0) satisfies y ≤ 40.

Region represented by x + y ≥ 10:

The line x + y = 10 meets the coordinate axes at A2 (10, 0) and B3 (0, 10) respectively. 

Region not containing origin represents x + y ≥ 10 as (0, 0) does not satisfy x + y ≥ 10.

Shaded region A3 A2 P Q B2 B3 represents feasible region.

Point P(40, 20) is obtained by solving x = 40 and x + y = 60

Point Q(20, 40) is obtained by solving y = 40 and x + y = 60

The value of Z = 3x + 4y + 370 at

A3(10, 0) = 3(10) + 4(0) + 370 = 400

A2(40, 0) = 3(40) + 4(0) + 370 = 490

P(40, 20) = 3(40) + 4(20) + 370 = 570

Q(20, 40) = 3(20) + 4(40) + 370 = 590

B2(0, 40) = 3(0) + 4(40) + 370 = 530

B3(0, 10) = 3(0) + 4(10) + 370 = 410

Hence, minimum value of Z = 400 at x = 10, y = 0

So, 

From A -> P = 10 packets

From A -> Q = 0 packets

From A -> R = 50 packets

From B -> P = 30 packets

From B -> Q = 40 packets

From B -> R = 0 packets

Therefore, minimum cost = Rs 400

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