# Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.1 | Set 1

**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:

GadgetFoundryMachine-shopProfitA 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 :

ProductMachine hoursLabour hoursProfitA 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:**

ProductMan HoursMaximum demandProfitA 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 Rs.