NCERT Solution Class 12 Chapter 12 - Linear Programming Exercise 12.2

NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.2 – Free PDF Download

Exercise 12.2 of NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming is based on the following topics:

1. Different Types of Linear Programming Problems
   1. Manufacturing problems
   2. Diet problems
   3. Transportation problems

Solve all the problems of this exercise to get thorough with the concepts and topics covered in the chapter.

NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming Exercise 12.2

Download PDF

carouselExampleControls112

Previous





Next

Access Other Exercises of Class 12 Maths Chapter 12

Exercise 12.1 Solutions 10 Questions

Miscellaneous Exercise on Chapter 12 Solutions 10 Questions

Access Answers to NCERT Class 12 Maths Chapter 12 Exercise 12.2

1. 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/kg and Food Q costs Rs 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B, while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

Solution:

Let the mixture contain x kg of food P and y kg of food Q.

Hence, x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below:

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Hence, the constraints are

3x + 4y ≥ 8

5x + 2y ≥ 11

The total cost of purchasing food is, Z = 60x + 80y

So, the mathematical formulation of the given problem can be written as

Minimise Z = 60x + 80y (i)

Now, subject to the constraints,

3x + 4y ≥ 8 … (2)

5x + 2y ≥ 11 … (3)

x, y ≥ 0 … (4)

The feasible region determined by the system of constraints is given below:

Clearly, we can see that the feasible region is unbounded

A(8/3, 0), B(2, 1/2) and C(0, 11/2)

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

Here, the feasible region is unbounded; therefore, Rs. 160 may or may not be the minimum value of Z.

For this purpose, we graph the inequality, 60x + 80y < 160 or 3x + 4y < 8, and check whether the resulting half-plane has points in common with the feasible region or not,

Here, it can be seen that the feasible region has no common point with 3x + 4y < 8

Hence, at the line segment joining the points (8/3, 0) and (2, 1/2), the minimum cost of the mixture will be Rs. 160.

2. One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingredients used in making the cakes.

Solution:

Let the first kind of cake be x and the second kind of cake be y. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as shown below:

So, 200x + 100y ≤ 5000

2x + y ≤ 50

25x + 50y ≤ 1000

x + 2y ≤ 40

The total number of cakes Z that can be made are

Z = x + y

The mathematical formulation of the given problem can be written as

Maximise Z = x + y (i)

Here, subject to the constraints,

2x + y ≤ 50 (ii)

x + 2y ≤ 40 (iii)

x, y ≥ 0 (iv)

The feasible region determined by the system of constraints is as given below:

A (25, 0), B (20, 10), O (0, 0) and C (0, 20) are the corner points

The values of Z at these corner points are as given below:

Hence, the maximum numbers of cakes that can be made are 30 (20 cakes of one kind and 10 cakes of another kind).

3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making, while a cricket bat takes 3 hours of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(ii) What number of rackets and bats must be made if the factory is to work at full capacity?

(ii) If the profit on a racket and on a bat is Rs. 20 and Rs. 10, respectively, find the maximum profit of the factory when it works at full capacity.

Solution:

Let x and y be the number of rackets and the number of bats to be made.

Given that the machine time is not available for more than 42 hours

Hence, 1.5x + 3y ≤ 42 ……………. (i)

Also, given that the craftsman’s time is not available for more than 24 hours

Hence, 3x + y ≤ 24 ………… (ii)

The factory is to work at full capacity. Hence,

1.5x + 3y = 42

3x + y = 24

On solving these equations, we get

x = 4 and y = 12

Therefore, 4 rackets and 12 bats must be made.

(i) The given information can be compiled in a table as given below;

1.5x + 3y ≤ 42

3x + y ≤ 24

x, y ≥ 0

Since the profit on a racket is Rs. 20 and Rs. 10,

Z = 20x + 10y

The mathematical formulation of the given problem can be written as

Maximise Z = 20x + 10y ………….. (i)

Subject to the constraints,

1.5x + 3y ≤ 42 …………. (ii)

3x + y ≤ 24 …………….. (iii)

x, y ≥ 0 ………………… (iv)

The feasible region determined by the system of constraints is as given below:

A(8, 0), B(4, 12), C(0, 14) and O(0, 0) are the corner points, respectively.

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

Therefore, the maximum profit of the factory when it works to its full capacity is Rs. 200.

4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs. 17.50 per package on nuts and Rs. 7.00 per package on bolts. How many packages of nuts and bolts should be produced each day so as to maximise his profit if he operates his machines for at the most 12 hours a day?

Solution:

Let the manufacturer produce x package of nuts and y package of bolts. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below:

The profit on a package of nuts is Rs. 17.50, and on a package of bolts is Rs. 7

Hence, the constraints are

x + 3y ≤ 12

3x + y ≤ 12

Then, total profit, Z = 17.5x + 7y

The mathematical formulation of the given problem can be written as follows

Maximise Z = 17.5x + 7y …………. (1)

Subject to the constraints,

x + 3y ≤ 12 …………. (2)

3x + y ≤ 12 ………… (3)

x, y ≥ 0 …………….. (4)

The feasible region determined by the system of constraints is given below:

A(4, 0), B(3, 3) and C(0, 4) are the corner points

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

Therefore, Rs. 73.50 at (3, 3) is the maximum value of Z.

Hence, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs. 73.50.

5. A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand-operated one. It takes 4 minutes on the automatic and 6 minutes on 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 machines 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 Rs. 7 and screw B at a profit of Rs. 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

Solution:

On each day, let the factory manufacture x screws of type A and y screws of type B.

Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below:

The profit on a package of screws A is Rs. 7, and on the package screws B is Rs. 10

Hence, the constraints are

4x + 6y ≤ 240

6x + 3y ≤ 240

Total profit, Z = 7x + 10y

The mathematical formulation of the given problem can be written as

Maximise Z = 7x + 10y …………. (i)

Subject to the constraints,

4x + 6y ≤ 240 …………. (ii)

6x + 3y ≤ 240 …………. (iii)

x, y ≥ 0 ……………… (iv)

The feasible region determined by the system of constraints is given below:

A(40, 0), B(30, 20) and C(0, 40) are the corner points

The value of Z at these corner points is given below:

The maximum value of Z is 410 at (30, 20).

Hence, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs. 410.

6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. 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 at most 12 hours. The profit from the sale of a lamp is Rs. 5, and that from a shade is Rs. 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?

Solution:

Let the cottage industry manufacture x pedestal lamps and y wooden shades, respectively

Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below:

The profit on a lamp is Rs. 5 and on the shades is Rs. 3. Hence, the constraints are

2x + y ≤ 12

3x + 2y ≤ 20

Total profit, Z = 5x + 3y …………….. (i)

Subject to the constraints,

2x + y ≤ 12 …………. (ii)

3x + 2y ≤ 20 ………… (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the system of constraints is given below:

A(6, 0), B(4, 4) and  (0, 10) are the corner points

The value of Z at these corner points is given below:

The maximum value of Z is 32 at point (4, 4).

Therefore, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.

7. 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 and 20 minutes available for cutting and 4 hours for assembling. The profit is Rs. 5 each for type A and Rs. 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

Solution:

Let the company manufacture x souvenirs of type A and y souvenirs of type B, respectively

Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below:

The profit on type A souvenirs is Rs. 5 and on type B souvenirs is Rs. 6. Hence, the constraints are

5x + 8y ≤ 200

10x + 8y ≤ 240 i.e.,

5x + 4y ≤ 120

Total profit, Z = 5x + 6y

The mathematical formulation of the given problem can be written as

Maximise Z = 5x + 6y …………… (i)

Subject to the constraints,

5x + 8y ≤ 200 ……………. (ii)

5x + 4y ≤ 120 ………….. (iii)

x, y ≥ 0 ………….. (iv)

The feasible region determined by the system of constraints is given below:

A(24, 0), B(8, 20) and C(0, 25) are the corner points

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

The maximum value of Z is 200 at (8, 20).

Hence, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs 160.

8. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs. 25000 and Rs. 40000, respectively. He estimates that the total monthly demand for computers will not exceed 250 units. Determine the number of units of each type of computer that the merchant should stock to get maximum profit if he does not want to invest more than Rs. 70 lakhs and if his profit on the desktop model is Rs. 4500 and on the portable model is Rs. 5000.

Solution:

Let the merchant stock x desktop models and y portable models, respectively.

Hence,

x ≥ 0 and y ≥ 0

Given that the cost of the desktop model is Rs. 25000 and of a portable model is Rs. 40000.

However, the merchant can invest a maximum of Rs. 70 lakhs

Hence, 25000x + 40000y ≤ 7000000

5x + 8y ≤ 1400

The monthly demand of computers will not exceed 250 units.

Hence, x + y ≤ 250

The profit on a desktop model is Rs. 4500, and the profit on a portable model is Rs. 5000

Total profit, Z = 4500x + 5000y

Therefore, the mathematical formulation of the given problem is

Maximum Z = 4500x + 5000y ………… (i)

Subject to the constraints,

5x + 8y ≤ 1400 ………… (ii)

x + y ≤ 250 ………….. (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the system of constraints is given below:

A(250, 0), B(200, 50) and C(0, 175) are the corner points.

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

The maximum value of Z is 1150000 at (200, 50).

Therefore, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs. 1150000.

9. 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 of food, 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. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.

Solution:

Let the diet contain x units of food F₁ and y units of food F₂. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below:

The cost of food F₁ is Rs 4 per unit, and of food F₂ is Rs. 6 per unit

Hence, the constraints are

3x + 6y ≥ 80

4x + 3y ≥ 100

x, y ≥ 0

The total cost of the diet, Z = 4x + 6y

The mathematical formulation of the given problem can be written as

Minimise Z = 4x + 6y …………… (i)

Subject to the constraints,

3x + 6y ≥ 80 ………… (ii)

4x + 3y ≥ 100 ………. (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the constraints is given below:

We can see that the feasible region is unbounded.

A(80/3, 0), B(24, 4/3), and C (0, 100/3) are the corner points

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

Here, the feasible region is unbounded, so 104 may or not be the minimum value of Z.

For this purpose, we draw a graph of the inequality, 4x + 6y < 104 or 2x + 3y < 52, and check whether the resulting half-plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with 2x + 3y < 52.

Hence, the minimum cost of the mixture will be Rs. 104.

10. There are two types of fertilisers F₁ and F₂. F₁ consists of 10% nitrogen and 6% phosphoric acid, and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁ costs Rs. 6/kg and F₂ costs Rs. 5/kg, determine how much of each type of fertiliser should be used, so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Solution:

Let the farmer buy x kg of fertiliser F₁ and y kg of fertiliser F₂. Hence,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as given below:

F₁ consists of 10% nitrogen, and F₂ consists of 5% nitrogen.

However, the farmer requires at least 14 kg of nitrogen

So, 10% of x + 5% of y ≥ 14

x / 10 + y / 20 ≥ 14

By LCM, we get

2x + y ≥ 280

F₁ consists of 6% phosphoric acid, and F₂ consists of 10% phosphoric acid.

However, the farmer requires at least 14 kg of phosphoric acid

So, 6% of x + 10 % of y ≥ 14

6x / 100 + 10y / 100 ≥ 14

3x + 5y ≥ 700

The total cost of fertilisers, Z = 6x + 5y

The mathematical formulation of the given problem can be written as

Minimize Z = 6x + 5y ………….. (i)

Subject to the constraints,

2x + y ≥ 280 ……… (ii)

3x + 5y ≥ 700 ………. (iii)

x, y ≥ 0 …………. (iv)

The feasible region determined by the system of constraints is given below:

Here, we can see that the feasible region is unbounded.

A(700/3, 0), B(100, 80) and C(0, 280) are the corner points

The values of Z at these points are given below:

Here, the feasible region is unbounded; hence, 1000 may or may not be the minimum value of Z.

For this purpose, we draw a graph of the inequality, 6x + 5y < 1000, and check whether the resulting half-plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with 6x + 5y < 1000.

Hence, 100 kg of fertiliser F₁ and 80 kg of fertiliser F₂ should be used to minimise the cost. The minimum cost is Rs. 1000.

11. The corner points of the feasible region are determined by the following system of linear inequalities:

2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is

(A) p = q

(B) p = 2q

(C) p = 3q

(D) q = 3p

Solution:

The maximum value of Z is unique

Here, it is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5)

Value of Z at (3, 4) = Value of Z at (0, 5)

p (3) + q (4) = p (0) + q (5)

3p + 4q = 5q

3p = 5q – 4q

3p = q or q = 3p

Therefore, the correct answer is option (D).

Also, explore –

NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12

NCERT Solutions