PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2

Question 1.
Reshma wishes to mix two types of food P and Q in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹ 60/kg and food Q costs ₹ 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 contains x kg of food P and y kg of food Q.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 1

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B.
Therefore, the constraints are 3x + 4y ≥ 8; 5x + 2y ≥ 11.
Total cost, Z of purchasing food is Z = 60x + 80 y
The mathematical formulation of the given prolem is

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 2

Minimise Z = 60x + 80y …………(i)
subject to the constraints,
3x + 4y ≥ 8 ……………(ii)
5x + 2y ≥ 1 …………….(iii)
x,y ≥ 0 …………..(iv)
The feasible region determined by the system of constraints is as follows.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (\(\frac{8}{3}\), 0), (2, \(\frac{1}{2}\)) and C (0, \(\frac{11}{2}\)).
The value of Z at these corner points are as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 3

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this, we graph the inequality, 60x + 80 y < 160 or 3x + 4y < 8 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 4y< 8.
Therefore, the minimum cost of the mixture will be ₹ 160 at the line segment joining the points (\(\frac{8}{3}\), 0) and (2, \(\frac{1}{2}\)).

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 2.
One kind of cake requires 200g of flour and 25g of fat and another kind of cake requireds 100g of flour and 50g of fat. Find the maximum number of cakes which can he made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
Solution.
Let there be x cakes of first kind and y cakes of second kind.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 4

∴ 200x +100y ≤ 5000
⇒ 2x + y ≤ 50
25x + 50 y ≤ 1000
⇒ x + 2y ≤ 40 Total number of cakes, Z, that can be made are, Z = x + y
The mathematical formulation of the given problem is
Maximise Z = x + y ……………(i)
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 follows

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 5

The corner points are A(25, 0), B(20, 10), C(0, 20) and 0(0, 0). The values of Z at these comer points are as follows

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 6

Thus, the maximum number of cakes that can be made are 30 (20 of one kind and 10 of the other kind).

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 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 hour 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.
(i) 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 ₹ 20 and ₹ 10 respectively, find the maximum profit of the factory when it works at full capacity.
Solution.
(i) Let the number of rackets and the number of bats to be made by x and y respectively.
The machine time is not available for more than 42 hours.
∴ 1.5x + 3y ≤ 42 …………(i)
The craftsman’s time is not available for more than 42 hours.
∴ 3x + y ≤ 24 ……………(ii)
The factory is to work at full capacity.
Therefore, 1.5x + 3y = 42;
3x + y = 24
On solving these equations, we get x = 4 and y = 12
Thus, 4 rackets and 12 bats must be made.

(ii) The given information can be compiled in a tables as follows

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 7

∴ 1.5x + 3y ≤ 42;
3x + y ≤ 24;
x, y ≥ 0
The profit on a racket is ₹ 20 and on a bat is ₹ 10.
∴ Z = 20x +10y
The mathematical formulation of the given problem is
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 follows

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 8

The corner points are A (8, 0) B (4, 12), C (0, 14) and O (0, 0).
The values of Z at these comer points are as follows

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 9

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

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 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 holts. He earns a profit, of ₹ 17.50 per package on nuts and ₹ 7.00 per package on bolts. How many packages of each 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 packages of nuts and y packages of bolts.
Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 10

The profit on a package of nuts if ₹ 17.50 and on a package of bolts is ₹ 7.
Therefore, the constraints are x + 3y ≤ 12; 3x + y ≤ 12.
Total profit Z = 17.5x + 7y
The mathematical formulation of the given problem is
Maximise Z = 17.5x + 7y ……………(i)
subject to the constraints, x + 3y ≤12 ……….(ii)
3x + y ≤ 12………….(iii)
x, y ≥ 0 …………(iv)
The feasible region determined by the system of constraints is as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 11

The corner points are A(4, 0), B(3, 3) and C(0, 4).
The values of Z at these corner points are as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 12

The maximum value of Z is ₹ 73.50 at (3, 3).
Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit ₹ 73.50.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 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. 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 automatic and 3 minutes on the hand operated machines to manufacture a package of screws B.

Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ₹ 7 and screws B at a profit of ₹ 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.
Let the factory manufacture x screws of type A and y screws of type B on each day.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 13

The profit on a package of screws A is ₹ 7 and on the package of screws B is ₹ 10.
Therefore the constraints are 4x + 6y ≤ 240; 6x + 3y ≤ 240
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem is
Maximise Z = 7x+10y …………..(i)
subjectto the constraints, 4x + 6y ≤ 240 …………..(ii)
6x + 3y ≤ 240 …………..(iii)
x, y ≥ 0 …………….(iv)
The feasible region determined by the system of constraints is shown in previous page graph.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 14

The comer points are A (40, 0), B (30, 20) and C (0, 40).
The values of Z at these comer points are as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 15

The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit ₹ 410.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 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 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 the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is ₹ 5 and that from a shade is ₹ 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.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 16

The profit on a lamp is ₹ 5 and on the shades is ₹ 3.
Therefore, the constraints are 2x + y ≤ 12; 3x + 2y ≤ 20
Total profit, Z = 5x + 3y ,
The mathematical formulation of the given problem is
Maximise 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 as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 17

The maximum value of Z at these comer points are as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 18

The maximum value of Z is 32 at (4, 4).
Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 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 20 minutes available for cutting and 4 hours for assembling. The profit is ₹ 5 each for type A and ₹ 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.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in the table as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 19

The profit on type A souvenirs is ₹ 5 and on type B souvenirs is ₹ 6.
Therefore, the constraints are
5x + 8y ≤ 200;
10x +8y ≤ 200 i.e., 5x +4y ≤ 120

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 20

The mathematical formulation of the given problem is
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 as follows.
The corner points are A(24, 0), B(8, 20) and C(0, 25).
The values of Z and these comer points are as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 21

The maximum value of Z is 200 at (8, 20).
Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of ₹ 160.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 8.
A merchant plans to sell two types of personal computers — a desktop model and a portable model that will cost ₹ 25000 and ₹ 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹ 70 lakhs and if his profit on the desktop model is₹ 4500 and on portable model is ₹ 5000.
Solution.
Let the merchant stock x desktop models and y portable models.
Therefore, x ≥ 0 and y ≥ 0
The cost of a desktop model is ₹ 25000 and of a portable model ₹ 40000.
However, the merchant can invest a maximum of ₹ 70 lakhs.
∴ 25000x + 40000y ≤ 7000000
⇒ 5x + 8y ≤ 1400
The monthly demand of computers will not exceed 250 units.
∴ x + y ≤ 250
The profit on a desktop model is ₹ 4500 and the profit on a portable model is ₹ 5000.
Total profit, Z = 4500x + 5000y
Thus, the mathematical formulation of the given problem is
Maximise 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 as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 22

The corner points are A (250, 0), B (200, 50) and C (0, 175).
The values of Z at these corner points are as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 23

The maximum value of Z is 1150000 at (200, 50).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of ₹ 1150000.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 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 ₹ 4 per unit food and F2 costs ₹ 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 diet that consists of 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.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 24

The values of Z at these comer points are as follows.
The cost of food F1is 4 per unit and of food F2 is 6 per unit.
Therefore, the constraints are 3x + 6y ≥80; 4x + 3y ≥100; x, y ≥ 0
Total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem is
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 as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 25

It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A (\(\frac{8}{3}\), o) B (2, \(\frac{1}{2}\)) and C(o, \(\frac{11}{2}\))
The comer points are A (\(\frac{80}{3}\), B (24, \(\frac{4}{3}\)) and C (o, \(\frac{100}{3}\))

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

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 26

As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.
For this, 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.
It can be seen that the feasible region has no common point with 2x +3y < 52.
Therefore, the minimum cost of the mixture will be ₹ 104.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 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 ₹ 6per kg and F2 costs ₹ 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.
Therefore, x ≥ 0 and y ≥ 0.
The given information can be compiled in a table as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 27

F1 consists of 10% nitrogen and F2 consists of 5% nitrogen.
However, the farmer requires adeast 14 kg of nitrogen.
∴ 10% of x + 5% of y ≥ 14
⇒ \(\frac{x}{10}+\frac{y}{20}\) ≥ 14
⇒ 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.
∴ 6 % of x +10 % of y ≥ 14
⇒ \(\frac{6 x}{100}+\frac{10 y}{100}\) ≥ 14
⇒ 3x + 56y ≥ 700
Total cost of fertilisers, Z = 6x + 5y
The mathematical formulation of the given problem is
Minimise 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 as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 28

It can be seen that the feasible region is unbounded.
The comer points are A (\(\frac{700}{3}\), o) B (100, 80) and C (0, 280).
The values of Z of these points are as follows.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2 29

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.
For this, 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.
It can be seen that the feasible region has no common point with 6x + 5y < 1000
Therefore, 100 kg of fertiliser F1 and 80 kg of fertlizer F2 should be used to minimise the cost.
The minimum cost is ₹ 1000.

PSEB 12th Class Maths Solutions Chapter 12 Linear Programming Ex 12.2

Question 11.
The comer points of the feasible region determined by the following system of liner 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.
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
⇒ q = 3P
Hence, the correct answer is (D).

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