Here we provide RD Sharma Class 12 Ex 30.1 Solutions Chapter 30 Linear Programming for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 30.1 Solutions Chapter 30 Linear Programming book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 30 |

Exercise | 30.1 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 30.1 Solutions Chapter 30 Linear Programming**

**Question 1. **A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man – hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:

Gadget | Foundry | Machine-shop |

A | 10 | 5 |

B | 6 | 4 |

Firm’s capacity per week | 1000 | 60 |

### The profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as LPP.

**Answer: **

The given data may be put in the following tabular form:

Gadget | Foundry | Machine-shop | Profit |

A | 10 | 5 | Rs. 30 |

B | 6 | 4 | Rs. 20 |

Firm’s capacity per week | 1000 | 600 |

Let the required weekly production of gadgets A and B be x and y respectively.

Given that, profit on each gadget A is Rs 30 and gadget B is Rs 20.

Profit on x gadget of type A = 30x

Profit on y gadget of Type B = 20y

Let Z denote the total profit, so

Z = 30x + 20y

Given, production of one gadget A requires 10 hours per week for foundry and gadget B requires 6 hours per week for foundry.

So, x units of gadget A requires 10x hours per week and y units of gadget B requires 6y hours per week. But the maximum capacity of foundry per week is 1000 hours. So,

10x + 6y <= 1000 (First constraint.)

Given, production of one unit gadget A requires 5 hours per week of machine shop and production off one unit of gadget B requires 4 hours per week of machine shop.

So, x units of gadget A requires 5x hours per week and y units of gadget B requires 4y hours per week, but the maximum capacity of machine shop is 600 hours per week.

So, 5x + 4y <= 600 (Second constraint.)

Hence, mathematical formulation of LPP is:

Find x and y which maximize Z = 30x + 20y

Subject to constraints,

10x + 6y <= 1000

5x + 4y <=600

And, x, y >=0 (Since production cannot be less than 0.)

**Question 2: **A company is making two products A and B. The cost of producing one unit of products A and B are Rs 60 and Rs 80 respectively. As per the agreement, company has to supply at least 200 units of product B to its regular customers. One unit of product A requires one machine hour whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirements. Formulate this problem as LPP.

**Answer:**

The given information can be written in tabular form as follows :

Product | Machine hours | Labour hours | Profit |

A | 1 | 1 | Rs 60 |

B | – | 1 | Rs 80 |

Total Capacity | 400 for A | 500 |

Minimum supply of product B is 200 units.

Let the production of product A be x units and production of product B be y units.

Profit on one unit of product A = Rs 60

Profit on x units of product A = Rs 60y

Profit on one unit of product B = Rs 80

Profit on y units of product B = Rs 80y

Let Z denote the total profit. So,

Z = 60x + 80y

Given, minimum supply of product B is 200.

So, y >= 200 (First constraint)

Given that the production of one unit of product A requires 1 hour of machine hours, so x units of product A requires x hours , but given total machine hours available for product A is 400 hours. So,

x <= 400 (Second constraint)

Given, each unit of product A and B requires one hour of labour hour, so x units of product A require x hour and y unit of product B require y hour of labour hours, but total labour hours available are 500. So,

x + y <= 500 (Third constraint)

Hence, mathematical formulation of LPP is,

Find x and y which minimize Z = 60x + 80y

Subject to constraints,

y >= 200,

x <= 400,

x + y <= 500

x, y >=0 (Since production cannot be less than 0.)

**Question 3: **A firm produces 3 products A, B, and c. The profit are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product:

Machine | Products | ||

A | B | C | |

M1 | 4 | 3 | 5 |

M2 | 2 | 2 | 4 |

### Machines M1 and M2 have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 A’s, 200 B’s and 50 C’s but not more than 150 A’s. Set up a LPP to maximize the profit.

**Answer:**

The given information can be written in tabular form as follows :

Product | Machine(M1) | Machine(M2) | Profit |

A | 4 | 2 | 3 |

B | 3 | 2 | 2 |

C | 5 | 4 | 4 |

Maximum capacity | 2000 | 2500 |

Let required production of product A, B and C be x, y and z units respectively.

Given, profit on one unit of product A, B and C are Rs 3, Rs. 2 and Rs 4 respectively

So, profit on x unit of A, y unit of B and z unit of C are given by Rs 3x, Rs. 2y and Rs 4z respectively.

Let U be the total profit, so

U = 3x + 2y + 4z

Given, one unit of product A, B and C requires 4, 3 and 5 minutes on machine M1. So, x units of product A, y units of product B and z units of product C requires 4x, 3y and 5z minutes on machine M1. Therefore,

4x + 3y + 5z <= 2000 (First constraint)

Given, one unit of product A, B and C requires 2, 2 and 4 minutes on machine M1. So, x units of product A, y units of product B and z units of product C requires 2x, 2y and 4z minutes on machine M2. Therefore,

2x + 2y + 4z <= 2500 (Second constraint)

Also, given that the firm must manufacture 100 A’s, 200 B’s and 50 C’s but not more than 150 A’s. So,

100 <= x <= 150

y >=200

z >=50

Hence the mathematical formulation of LPP is:

Find x, y and z which minimize U = 3x + 2y + 4z

Subject to constraints,

4x + 3y + 5z <= 2000

2x + 2y + 4z <= 2500

100 <= x <= 150

y >=200

z >=50

x, y, z >=0

**Question 4: **A firm manufactures two types of products A and B and sells them at a profit of Rs 2 on type A and Rs 3 on type B. Each product is processed on two machines M1 and M2. Type A requires one minute of processing time on M1 and two minutes of M2; type B requires one minute on M1 and one minute on M2. The machine M1 is available for not more than 6 hours 40 minutes while machine M2 is available for 10 hours during any working day. Formulate this problem as LPP.

**Answer:**

The given information can be written in tabular form as follows :

Product | M1 | M2 | Profit |

A | 1 | 2 | 2 |

B | 1 | 1 | 3 |

Capacity | 6 hours 40 min. = 400 min | 10 hours = 600 min. |

Let the required production of product A be x units and product B be y units.

Profit on one unit of product A = Rs 2

Profit on x units of product A = Rs 2x

Profit on one unit of product B = Rs 3

Profit on y units of product A = Rs 3y

Let the total profit be Z, so

Z = 2x + 3y

On machine M1,

Production of one unit of product A requires 1 minute.

Production of x units of product A requires x minute.

Production of one unit of product B requires 1 minute.

Production of y units of product B requires y minute.

But total time available on machine M1 is 600 minutes.

So, x + y <=400 (First constraint)

On machine M2,

Production of one unit of product A requires 2 minute.

Production of x units of product A requires 2x minute.

Production of one unit of product B requires 1 minute.

Production of y units of product B requires y minute.

But total time available on machine M2 is 600 minutes.

So, 2x + y <=600 (Second constraint)

Hence the mathematical formulation of LPP is:

Find x, y and z which maximize Z = 2x + 3y

Subject to constraints,

x + y <=400

2x + y <=600

x, y >=0 (Production cannot be less than zero.)

**Question 5: **A rubber company is engaged in producing three types of tyres A, B, and C, Each type requires processing n two plants, Plant 1 and Plant 2. The capacities of the two plants, in number of tyres per day, are as follows:

Plant | A | B | C |

1 | 50 | 100 | 100 |

2 | 60 | 60 | 200 |

### The monthly demand for tyre A, B and C is 2500, 3000 and 7000 respectively. If plant 1 costs R2 2500 per day, and plant 2 costs Rs 3500 per day to operate, how many days should each be run per month to minimize cost while meeting the demand? Formulate this problem as LPP.

**Answer:**

The given information can be written in tabular form as follows :

Plant | A | B | C | Cost |

1 | 50 | 100 | 100 | 2500 |

2 | 60 | 60 | 200 | 3500 |

Monthly demand | 2500 | 3000 | 7000 |

Let plant 1 requires x days and plant 2 requires y days per month to minimize cost.

Given, plant 1 and plant 2 requires Rs 2500 per day and Rs 3500 per day respectively.

So, cost to run plant 1 and 2 is Rs 2500x and Rs 3500y per month.

Let Z be the total cost per month, so

Z = 2500x + 3500y

Given production of tyre A from plant 1 and 2 per day is 50 and 60 respectively. So, production of tyre A from plant 1 and 2 per month will be 50x and 60y respectively. But the maximum demand of tyre A is 2500 per month. So,

50x + 60y >= 2500 (First constraint)

Given production of tyre B from plant 1 and 2 per day is 100 and 60 respectively. So, production of tyre B from plant 1 and 2 per month will be 100x and 60y respectively. But the maximum demand of tyre B is 3000 per month. So,

100x + 60y >= 3000 (Second constraint)

Given production of tyre C from plant 1 and 2 per day is 100 and 200 respectively. So, production of tyre C from plant 1 and 2 per month will be 100x and 200y respectively. But the maximum demand of tyre A is 7000 per month. So,

100x + 200y >= 7000 (Third constraint)

Hence the mathematical formulation of LPP is:

Find x and y which minimize Z = 2500x + 3500y

Subject to constraints,

50x + 60y >= 2500

100x + 60y >= 3000

100x + 200y >= 7000

x, y >= 0 (Since number of days cannot be less than zero)

**Question 6: **A company sells two different products A and B. The two products are produced in a common production process and are sold in two different markets. The production process has a total capacity of 45000 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 units of A that can be sold is 7000 and that of B 10000. If the profit is Rs 60 per unit for the product A and Rs 40 for product B, how many units of each product should be sold to maximize profit? Formulate this problem as LPP.

**Answer:**

Product | Man Hours | Maximum demand | Profit |

A | 5 | 7000 | 60 |

B | 3 | 10000 | 40 |

Total Capacity | 45000 |

Let the required production of product A be x units and product B be y units.

Profit on one unit of product A = Rs 60

Profit on x units of product A = Rs 60x

Profit on one unit of product B = Rs 40

Profit on y units of product A = Rs 40y

Let the total profit be Z, so

Z = 60x + 40y

Production of one unit of product A requires 5 hours.

Production of x units of product A requires 5x hours.

Production of one unit of product B requires 3 hours.

Production of y units of product B requires 3y hours.

But the total man hours available are 45000 hours, so

5x + 3y <= 450000 (First constraint)

Given, maximum demand for product A is 7000, so

x <= 7000 (Second constraint)

Given, maximum demand for product B is 10000, so

y <= 10000 (Third constraint)

Hence the mathematical formulation of LPP is:

Find x and y which minimize Z = 2500x + 3500y

Subject to constraints,

5x + 3y <= 450000

x <= 7000

y <= 10000

x, y >= 0 (Since production cannot be less than zero)

**Question 7: **To maintain his health a person must fulfill certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients – calcium, protein and calories and the person’s diet consists of only two food items , 1 and 2, whose price and nutrient contents are shown in below table:

Food 1(per lb) | Food 2(per lb) | Minimum daily requirement for the nutrient | |

Calcium | 10 | 5 | 20 |

Protein | 5 | 4 | 20 |

Calories | 2 | 6 | 13 |

Price (Rs) | 60 | 100 |

### What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this problem as LPP.

**Answer:**

Let x and y be the packets of 25 gm of Food 1 and Food 2 purchased. Let Z be the price paid. Obviously we have to minimize the price.

Take a mass balance on the nutrients from Food 1 and 2,

Calcium: 10x + 4y >= 20, i.e., 5x + 2y >= 10 (Equation 1)

Protein: 5x + 5y >= 20, i.e., x + y >= 20 (Equation 2)

Calories: 2x + 6y >=13 (Equation 3)

These become the constraints for the cost function, Z to be minimized , i.e, 0.6x + y = Z, given cost of Food 1 is Rs 0.6 and Rs 1 per lb.

From equation 1, 2 and 3 we get points on the X & Y- axis as (0,5) & (2,0); (0,4) & (4,0); (0,13/6) & (6.5, 0).

Plotting these,

The smallest value of Z is 2.9 at the point (2.75, 1.25). We cannot say that the minimum value of Z is 2.9 as the feasible region is unbounded.

Therefore, we have to draw the graph of the inequality 0.6x + y < 2.9.

Plotting this to see if the resulting line has any plot common with the feasible region. Since there are no common points, this is the minimum value of the function Z and the mix is

Food 1 = 2.75 lb; Food 2 = 1.25 lb; Price = Rs 2.9

When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region.

Here the feasible region is the unbounded region A-B-C-D.

Computing the value of Z a t the corner points of the feasible region ABHG

Point | Corner Point | Value of Z = 0.6x + y |

A | 2 ,5 | 6.2 |

B | 0.67, 3.33 | 3.73 |

C | 2.75, 1.25 | 2.9 |

D | 6.5, 2.16 | 6.06 |

**Question 8: **A manufacturer can produce two products, A and B, during a given time period. Each of these products requires four different manufacturing operations: grinding, turning, assembling, and testing. The manufacturing requirements in hours per unit of products A and B are given below:

A | B | |

Grinding | 1 | 2 |

Turning | 3 | 1 |

Assembling | 6 | 3 |

Testing | 5 | 4 |

### The available capacities of these operations in hours for the given time period are: grinding 30; turning 60; assembling 200; testing 200. The contribution to profit is Rs 20 for each unit of A and Rs 30 for each unit of B. The firm can sell all that it produces at the prevailing market price. Determine the optimum amount of A and B to produce during the given time period. Formulate this problem as LPP.

**Answer:**

Product | Grinding | Turning | Assembling | Testing | Profit |

A | 1 | 3 | 6 | 5 | 2 |

B | 2 | 1 | 3 | 4 | 3 |

Maximum capacity | 30 hours | 60 hours | 200 hours | 200 hours |

Let the required production of product A be x units and product B be y units.

Profit on one unit of product A = Rs 2

Profit on x units of product A = Rs 2x

Profit on one unit of product B = Rs 3

Profit on y units of product A = Rs 3y

Let the total profit be Z, so

Z = 2x + 3y

Production of one unit of product A requires 1 hours of grinding.

Production of x units of product A requires x hours of grinding.

Production of one unit of product B requires 2 hours of grinding.

Production of y units of product B requires 2y hours of grinding.

But the total time available for grinding is 30 hours, so

x + 2y <= 30 (First constraint)

Production of one unit of product A requires 3 hours of turning.

Production of x units of product A requires 3x hours of turning..

Production of one unit of product B requires 1 hours of turning..

Production of y units of product B requires y hours of turning..

But the total time available for turning.is 60 hours, so

3x + y <= 60 (Second constraint)

Production of one unit of product A requires 6 hours of assembling.

Production of x units of product A requires 6x hours of assembling.

Production of one unit of product B requires 3 hours of assembling.

Production of y units of product B requires 3y hours of assembling.

But the total time available for assembling is 200 hours, so

6x + 3y <= 200 (Third constraint)

Production of one unit of product A requires 5 hours of testing.

Production of x units of product A requires 5x hours of testing.

Production of one unit of product B requires 4 hours of testing.

Production of y units of product B requires 4y hours of testing.

But the total time available for testing is 200 hours, so

5x + 4y <= 200 (Fourth constraint)

Hence the mathematical formulation of LPP is:

Find x and y which maximize Z = 2x + 3y

Subject to constraints,

x + 2y <= 30

3x + y <= 60

6x + 3y <= 200

5x + 4y <= 200

x, y >= 0 (Since production cannot be negative)

**Question 9: **Vitamins A and B are found in two different foods F1 and F2. One unit of food F1 contains 2 units of Vitamin A and 3 units of Vitamin B. One unit of food F2 contains 4 units of Vitamin A and 2 units of Vitamin B. One unit of food F1 and F2 cost Rs 50 and 25 respectively. The minimum daily requirements for a person of vitamin A and B is 40 and 50 units respectively. Assuming that anything in excess of daily minimum requirement of Vitamin a and B is not harmful, find the optimum mixture of food F1 and F2 at the minimum cost which meets the daily minimum requirement of vitamin A and B. Formulate this problem as LPP.

**Answer:**

Given information can be tabulated as below:

Foods | Vitamin A | Vitamin B | Cost |

F1 | 2 | 3 | 5 |

F2 | 4 | 2 | 2.5 |

Minimum daily requirement | 40 | 50 |

Let the required quantity of Food F1 be x units and quantity of food F2 be y units.

Given, cost of one unit of food F1 and F2 are Rs 5 and Rs 2.5 respectively. So, cost of x units of Food F1 and y units of food F2 are Rs 5x and 2.5y respectively.

Let Z be the total cost, so

Z = 5x + 2.5y

Given, one unit of food F1 and F2 contain 2 and 4 units of Vitamin A respectively, so x unit of food F1 and y units of food F2 contain 2x and 4y units of Vitamin A respectively, but minimum requirement of Vitamin A is 40 unit, so

2x + 4y >= 40 (First constraint)

Given, one unit of food F1 and F2 contain 3 and 2 units of Vitamin B respectively, so x unit of food F1 and y units of food F2 contain 3x and 2y units of Vitamin B respectively, but minimum requirement of Vitamin A is 50 unit, so

3x + 2y >= 50 (Second constraint)

Hence the mathematical formulation of LPP is:

Find x and y which maximize Z = 5x + 2.5y

Subject to constraints,

2x + 4y >= 40

3x + 2y >= 50

x, y>= 0 (Since requirement of food F1 and F2 cannot be less than zero)

**Question 10: **An automobile manufacturer makes automobiles and trucks in a factory that is divided into two shops. Shop A, which performs the basic assembly operation, must work 5 man-days on each truck but only 2 man-days on each automobile. Shop B, which performs the finishing operations, must work 3 man-days for each automobile or truck it produces. Because of men and machine limitations, shop A has 180 man-days per week available while shop B has 135 man-days per week. If the manufacturer makes a profit of Rs 30000 on each truck and Rs 2000 on each automobile, how many of each should he produce to maximize his profit? Shop A, which performs the basic assembly operation, must work 5 man-days on each truck but only 2 man-days on each automobile.

**Answer:**

Let the number of automobiles produced be x and let the number of trucks produced be y.

Let Z be the profit function to be maximized.

Z = 2000x + 30000y

The constraints are on the man hours worked

Shop A : 2x + 5y <= 180 (Equation 1)

Shop B : 3x + 3y <= 135 ((Equation 2)

Corner points can be obtained from

2x + 5y = 180 , i.e, x= 0; y = 36 & x = 90; y = 0

3x + 3y = 135 , i.e. , x =0; y = 45 & x = 45; y = 0

Solving Equation 1 and Equation 2 gives x = 15 & y = 30

Corner point | Value of Z = 2000x + 30000y |

0, 0 | 0 |

0, 36 | 1080000 |

15, 30 | 930000 |

45, 0 | 90000 |

0 automobiles and 36 trucks will give maximum profit of 1080000

**Question 11. **Two tailors A and B earn Rs 150 and Rs 200 per day respectively. A can stitch 6 shirts and 4 pants per day while B can stitch shirts and 4 pants per day. Form a linear programming problem to minimize the labour cost to produce at least 60 shirts and 32 pants.

**Answer:**

Taylor A | Taylor B | Limit | ||

Variable | x | + | y | |

Shirts | 6x | + | 10y | >=60 |

Pants | 4x | + | 4y | >=32 |

Earn Rs | 150 | + | 200 | Z |

The above LPP can be represented in the table above.

To minimize labour cost means to assume minimize the earnings, i.e, Min Z = 150x + 200y such that the constraints

x>=0; y >= 0 (at least 1 shirt and pant is required)

6x + 10y >= 60 (require at least 60 shirts)

4x + 4y >= 32 (require at least 32 pants)

Solving the above inequalities a s equations we get, x = 5 and y = 3

Other corner points obtained are (0, 6) & (10, 0), (0, 8) & (8, 0)

The feasible region is the open unbounded region A-E-D.

Point E(5, 3) may not be the minimal value. So, plot 150x + 200y < 1350 to see if there is a common region with A-E-D.

The green line has no common point, therefore

Corner point | Value of Z = 150x + 200y |

0, 8 | 0 |

10, 0 | 1500 |

5, 3 | 1350 |

Stitching 5 shirts and 3 pants minimizes labour cost to Rs. 1350.

**Question 12. **An airline agrees to charter plane for a group. The group needs at least 160 first class seats and at least 300 tourist class seats. The sirline must use at least two of its model 314 planes which have 20 first class and 30 tourist class seats. The airline will also use some of its model 535 planes which have 20 first class seats and 60 tourist seats. Each flight of a model 314 plane costs the company Rs 1 lakh, and each flight of a model 535 plane costs Rs 1.5 lakh. How many of each type of plane should be used to minimize the flight cost? Formulate this as a LPP.

### Given information can be tabulated as below:

Plane | First class | Tourist class | Cost |

Model 314 | 20 | 30 | 100000 |

Model 335 | 20 | 60 | 150000 |

Requirement minimum | 160 seats | 300 seats |

**Answer:**

Model 314 | Model 535 | Limit | ||

Variable | x | + | y | |

F class | 20x | + | 20y | >= 160 |

T class | 30x | + | 60y | >= 300 |

Cost | x lakh | + | 1.5y lakh | Z |

The above LPP can be represented in the table above.

The flight cost is to be minimized, i.e, Min Z = x +1.5y such that the constraints

x >= 2 (at least 2 planes of model 314 must be used)

y >= 0 (at least 1 planes of model 535 must be used)

20x + 20y >= 160 (require at least 160F class seats)

30x + 60y >= 300 (require at least 300T class seats)

Solving the above inequalities as equations we get,

When x = 0, y = 8 and when y = 0, x = 8

When x = 0, y = 5 and when y = 0, x = 10

We get an unbounded region 8-E-10 as a feasible solution. Plotting the corner points and evaluating we have,

Corner point | Value of Z = x + 1.5y |

10, 0 | 10 |

0, 8 | 12 |

6, 2 | 9 |

Since we obtained an unbounded region as the feasible solution a plot of Z ( x + 1.5y < 9) is plotted.

since there are no common points , point E is the point that gives the minimum value.

Using 6 planes of model 314 & 2 of model 535 gives minimum cost of 9 lakh rupees.

**Question 13. **Amit’s mathematics teacher has given him three very long list of problems with the instruction to submit not more than 100 of them (correctly solved) for credit. The problem in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amt knows from experience that he requires on the average 3 minutes to solve a 5 points problem, 2 minutes to solve a 4 points problem, and 4 minutes to solve a 6 point problem. Because he has other subjects to worry about, he cannot afford to devote more than 3.5 hours altogether to his mathematics assignment. Moreover, the first two sets of problems involve numerical calculations and he knows that he cannot stand more than 2.5 hours work on this type of problem. Under these circumstances, how many problems in each of these categories shall he do in order to get maximum possible credit for his efforts? Formulate this as a LPP.

**Answer:**

Given information can be tabulated as below

Sets | Time requirement | Points |

1 | 3 | 5 |

2 | 2 | |

3 | 4 | 6 |

Time for all three sets = 3.5 hours Time for Set 1 ans Set 2 = 2.5 hours Maximum number of questions = 100 |

Let there be x, y, z questions from set 1, 2 and 3 respectively.

Given, each question from set 1, 2 and 3 earn 5,4 and 6 points respectively. So x questions of set 1, y questions of set 2 and z questions of set 3 earn 5x, 4y and 6z points.

Let total point credit be U

So, U = 5x + 4y + 6z

Given, each questions of set 1, 2 and 3 require 3, 2 and 4 minutes respectively. So, x questions of set 1, y questions of set 2 and z questions of set 3 require 3x, 2y and 4z minutes respectively but given that total time to devote in all three sets is 3.5 hours = 210 minutes and the first two sets is 2.5 hours = 150 minutes.

So,

3x + 2y + 4z <= 210 (First constraint)

3x + 2y < =150 (Second constraint)

Given, total number of questions cannot exceed 100.

So, x + y + z <= 100 (Third constraint)

Hence, mathematical formulation of LPP is

Find x and y which maximize U = 5x + 4y + 6z

Subject to constraint,

3x + 2y + 4z <= 210

3x + 2y < =150

x + y + z <= 100

x, y, z >= 0 (Since number of questions to solve from each set cannot be less than zero)

**Question 14. **A farmer has a 100 acre farm. He can sell the tomatoes, lettuce or radishes he can raise. The price he can obtain is Re 1 per kilogram for tomatoes, Rs 0.75 a head for lettuce and Rs 2 per kilogram for radishes. The average yield per acre is 2000 kgs for radishes , 3000 heads of lettuce and 1000 kilograms of radishes. Fertilizer is availableat Rs 0.05 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man days for tomatoes and radishes and 6 man days for lettuce. A total of 400 man days of labour are available at Rs 20 per man day. Formulate this problem as a LPP to maximize the farmer’s total profit.

**Answer:**

Given information can be tabulated as below:

Product | Yield | Cultivation | Price | Fertilizers |

Tomatoes | 2000 kg | 5 days | 1 | 100 kg |

Lettuce | 3000 kg | 6 days | 0.75 | 100 kg |

Radishes | 1000 kg | 5 days | 2 | 50 kg |

Average 2000 kg/per acre

Total land = 100 Acre

Cost of fertilizers = Rs 0.50 per kg

A total of 400 days of cultivation labour with Rs 20 per day

Let required quantity of field for tomatoes, lettuce and radishes be x, y and z acre respectively.

Given, costs of cultivation and harvesting of tomatoes, lettuce and radishes are 5 * 20 = Rs 100, 6 * 20 = Rs 120, 5 * 20 = Rs 100 respectively per acre. Cost of fertilizers for tomatoes, lettuce and radishes 100 * 0.05 = Rs 50, 100 * 0.50 = Rs 50 ans 50 * 0.50 = Rs 25 respectively per acre.

So, total costs of production of tomatoes, lettuce and radishes are Rs 100 + 50 = Rs 150x, Rs 120 + 50 = Rs 170y and Rs 100 + 25 = Rs 125z respectively. Total selling price of tomatoes, lettuce according to yield are 2000 * 1 = Rs 2000x, 3000 * 0.75 = Rs 2250y and 1000 * 2 = Rs 2000z respectively.

Let U be the total profit,

So,

U = (2000x – 150x) + (2250y – 170y) + (2000z – 125z)

U = 1850x + 2080y + 1875z

Given, farmer has 100 acre from

So, x + y + z <= 100 (First constraint)

Number of cultivation and harvesting days are 400

So, 5x + 6y + 5z <= 400 (Second constraint)

Hence, mathematical formulation of LPP is

Find x, y, z which maximize U = 11850x + 2080y + 1875z

Subject to constraint,

x + y + z <= 100

5x + 6y + 5z <= 400

x, y, z >=0 (Since cultivation cannot be less than zero)

**Question 15. **A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:

Product A | Product B | Weekly capacity | |

Department 1 | 3 | 2 | 130 |

Department 2 | 4 | 6 | 260 |

Selling price per unit | Rs 25 | Rs 30 | |

Labour price per unit | Rs 16 | Rs 20 | |

Raw material cost per unit | Rs 14 | Rs 4 |

### The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.

**Answer:**

Given information can be tabulated as below:

Product | Department 1 | Department 2 | Selling price | Labour cost | Raw material cost |

A | 3 | 4 | 25 | 16 | 4 |

B | 2 | 6 | 30 | 20 | 4 |

Capacity | 130 | 260 |

Let the required number of product A and B be x and y units respectively.

Given, labour cost and raw material cost of one unit of product A is Rs 16 and Rs 4, so total cost of product A is Rs 16 + Rs 4 = Rs 20

And given selling price of 1 unit of product A is Rs 25

So, profit on one unit of product A = Rs 25 – Rs 20 = Rs 5

Again, labour cost and raw material cost of one unit of product A is Rs 20 and Rs 4, so total cost of product A is Rs 20 + Rs 4 = Rs 24

And given selling price of 1 unit of product B is Rs 30

So, profit on one unit of product B = Rs 30 – Rs 24 = Rs 6

Hence, profits on x unit of product A and y units of product B are Rs 5x and Rs 6y respectively.

Let Z be the total profit, so Z = 5x + 6y

Given, production of one unit of product A and B need to process for 3 and 4 hours respectively in department 1, so production of x units of product A and y units of product B need to process for 3x and 4y hours respectively in department 1. But total capacity of Department 1 is 130 hours,

So, 3x + 2y <= 130 (First Constraint)

Given, production of one unit of product A and B need to process for 4 and 6 hours respectively in department 2, so production of x units of product A and y units of product B need to process for 4x and 6y hours respectively in department 2. But total capacity of Department 2 is 260 hours,

So, 4x + 6y <= 260 (Second Constraint)

Hence, mathematical formulation of LPP is

Find x, y, z which maximize Z = 5x + 6y

Subject to constraint,

3x + 2y <= 130

4x + 6y <= 260

x, y >= 0 (Since production cannot be less than zero)

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.