NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Maths
Chapter : 12
Chapter Name : Linear Programming

Q1 Solve the Linear Programming Problems graphically:
Maximise z = 3x + 4y

Subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0

Answer. The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0 is as follows.

The corner points of the feasible region are 0 (0, 0), A (4, 0), and B (0, 4). The values of z at these points
are as follows.

Therefore, the maximum value of Z is 16 at the point B (0 , 4).

Page : 513 , Block Name : Exercise 12.1

Q2 Solve the Linear Programming Problems graphically:
Minimise z = −3x + 4y

subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

Answer. The feasible region determined by the system of constraints,
x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, and y ≥ 0, as follows.

The corner points of the feasible region are O (O, O), A (4, O), B (2, 3), and C (O, 4).
The values of z at these corner points are as follows.

Therefore, the minimum value of Z is -12 at the point (4, 0).

Page : 514 , Block Name : Exercise 12.1

Q3 Solve the Linear Programming Problems graphically:
Maximise z = 5x + 3y

subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

The feasible region determined by the system of constraints, 3x + 5y ≤ 15 ,

5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

20 45
The corner points of the feasible region are O(0, 0), A(2, 0), B(0, 3), and C ( , )
19 19

The values of z at these corner points are as follows.

20 45
z is 19 at the point ( , )
19 19

Page : 514 , Block Name : Exercise 12.1

Q4 Solve the Linear Programming Problems graphically:
Minimise z = 3x + 5y

such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0

The feasible region determined by the system of constraints, x + 3y ≥ 3, x + y ≥ 2, and x,

v ≥ 0, is as follows.

It can be seen that the feasible region is unbounded.

As the feasible region is unbounded, therefore, 7 may or may not be the minimum value
For this, we draw the graph Of the inequality, 3x + 5y < 7, 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 + 5y < 7 Therefore,

The minimum value f z is 7at (
3 1
, )
2 2

Page : 514 , Block Name : Exercise 12.1

Q5 Solve the Linear Programming Problems graphically:
Maximise z = 3x + 2y

subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0

The feasible region determined by the constraints, x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, and

y ≥ 0, is as follows.

Page : 514 , Block Name : Exercise 12.1

Q6 Solve the Linear Programming Problems graphically:
Minimise z = x + 2y

subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0

The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0,≥ is as follows.

The comer points of the feasible region are A (6,0) and B (0,3). The values of Z at these corner points
are as follows.

It can be seen that the value of Z at points A and B is same. If we take any other point such as (2, 2) on
line x + 2y = 6, then Z = 6
Thus, the minimum value Of z occurs for more than 2 points .
Therefore, the value of Z Is minimum at every point on the line, x + 2y = 6

Page : 514 , Block Name : Exercise 12.1

Q7 Show that the minimum of Z occurs at more than two points.
Minimise and Maximise z = 5x + 10y

subject to x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x, y ≥ 0

The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥

0, x ≥ 0, and y ≥ 0, is as follows.

The corner points of the feasible region are A (60, O), B (120, O), C (60, 30), and C) (40, 20)
The values of Z at these corner points are as follows.

The minimum value of Z is 300 at (60, 0) and the maximum value Of Z is 600 at all the points on the
line segment joining (120, 0) and (60, 30).

Page : 514 , Block Name : Exercise 12.1

Q8 Show that the minimum of Z occurs at more than two points.
Minimise and Maximise z = x + 2y

subject to x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200; x, y ≥ 0

The feasible region determined by the constraints, x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤

200, x ≥ 0, and y ≥ 0, is as follows.

The corner points of the feasible region are A(O, 50), B(20, 40), C(50, 100), and D(0,
200). The values of Z at these corner points are as follows.

The maximum value of Z is 400 at (0, 200) and the minimum value of Z is 100 at all the
points on the line segment joining the points (0, 50) and (20, 40).

Page : 514 , Block Name : Exercise 12.1

Q9 Show that the minimum of Z occurs at more than two points.
Maximise z = −x + 2y, subject to the constraints

x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0

The feasible region determined by the constraints, x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, and y ≥ 0, is

as follows.

It can be seen that the feasible region is unbounded.
The values of Z at comer points A (6, 0), B (4, 1), and C (3, 2) are as follows.

As the feasible region is unbounded, therefore, Z 1 may or may not be the maximum
value.
For this, we graph the inequality, —x + 2y > 1, and check whether the resulting half
plane has points in common With the feasible region or not.
The resulting feasible region has points in common with the feasible region.
Therefore, Z = 1 is not the maximum value. Z has no maximum value.

Page : 514 , Block Name : Exercise 12.1

Q10 Show that the minimum of Z occurs at more than two points.
Z = x + y, subject to x − y ≤ −1, −x + y ≤ 0, x, y ≥ 0

Answer. x − y ≤ −1, −x + y ≤ 0, x, y ≥ 0, is as follows

Page : 514 , Block Name : Exercise 12.1

Q1 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 ot vitamin A and 11 units of vitamin 3. 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 3 while food Q
contains 4 units /kg of vitamin A and 2 units Jkg of vitamin 3. Determine the minimum cost of the
mixture?

Answer. Let the mixture contain 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.

The mixture must contain at least g units of vitamin A and 11 units of vitamin
Therefore, the constraints are
3x + 4y ≥ 8

5x + 2y ≥ 11

Total cost, Z , of purchasing food is, z = 60x + 80y

The mathematical formulation of the given problem is

Minimise Z = 60x + 80y … (1)

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 as follows.

It can be seen that the feasible region is unbounded.
8 1 11
A( , 0) , B (2, ) , and C (0, )
3 2 2

The values of z at these corner points are as follows.

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 + 80y < 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 Rs 160 at the line segment joining

8 1
( , 0) and (2, )
3 2

Page : 519 , Block Name : Exercise 12.2

Q2 One kind of cake requires 200g of our and 25g of fat, and another kind of cake requires 100g of
our and 50g of fat. Find the maximum number of cakes which can be made from 5 kg of our and 1 kg
of fat assuming that there is no shortage of the other ingredients used in making the cakes.

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 complied in a table as follows.

∴ 200x + 100y ≤ 5000

⇒ 2x + y ≤ 50

25x + 50y ≤ 1000

⇒ x + 2y ≤ 40

Total numbers of cakes, z , that can be made are, z = x + y

The mathematical formulation of the given problem is

Maximize Z = x + y … (1)

subject to the constraints,

2x + y ≤ 50

x + 2y ≤ 40

x, y ≥ 0

Page : 519 , Block Name : Exercise 12.2

Q3 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 pro t on a racket and on a bat is Rs 20 and Rs 10 respectively, nd the maximum pro t of the
factory when it works at full capacity.

(i) Let the number of rackets and the number of bats to be made be x and y

respectively.

The machine time is not available for more than 42 hours.

∴ 1.5x + 3y ≤ 42

The craftsman's time is not available for more than 24 hours.

∴ 3x + y ≤ 24

The factory is to work at full capacity. Therefore,

1.5x + y ≤ 24

3x + y = 24

On solving these equations, we obtain

x = 4 and y = 12

Thus, 4 rackets and 12 bats must be made.

(1) The given information can be complied in a table as follows.

∴ 1.5x + 3y ≤ 42

3x + y ≤ 24

x, y ≥ 0

The profit on a racket is Rs 20 and on a bat is Rs 10 .

∴ Z = 20x + 10y

The mathematical formulation of the given problem is

Maximize Z = 20x + 10y (1)

subject to the constraints,

1.5x + 3y ≤ 42 … (2)

3x + y ≤ 24 … (3)

x, y ≥ 0 … (4)

The feasible region determined by the system of constraints is as follows.

Page : 519 , Block Name : Exercise 12.2

Q4 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 pro t of Rs17.50 per package on nuts and Rs 7.00 per package on
bolts. How many packages of each should be produced each day so as to maximise his pro t, if he
operates his machines for at the most 12 hours a day?

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.

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

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 as follows.

Page : 519 , Block Name : Exercise 12.2

Q5 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
pro t of Rs 7 and screws B at a pro t 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 pro t? Determine the maximum pro t.

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.

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

Therefore, the constraints are

4x + 6y ≤ 240

6x + 3y ≤ 240

Total profit, Z = 7x + 10y

Maximize z = 7x + 10y … (1)

subject to the constraints,

4x + 6y ≤ 240

6x + 3y ≤ 240

6x + 3y ≤ 240 … (3)

x, y ≥ 0 … (4)

The feasible region determined by the system of constraints is

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 pro t of Rs 410.

Page : 519 , Block Name : Exercise 12.2

Q6 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 at the most 20
hours and the grinding/cutting machine for at the most 12 hours. The pro t 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 pro t?

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

x ≥ 0 and y ≥ 0

The given information can be compelled in a table as follows.

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

2x + y ≤ 12

3x + 2y ≤ 20

Total profit, z = 5x + 3y

The mathematical formulation of the given problem is

Maximize z = 5x + 3y … (1)

Maximize z = 5x + 3y … (1)

subject to the constraints,

2x + y ≤ 12 … (2)

3x + 2y ≤ 20

x, y ≥ 0 … (4)

Page : 520 , Block Name : Exercise 12.2

Q7 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 pro t 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 pro t?

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 complied in a table as follows.

The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6. Therefore, the

constraints are

5x + 8y ≤ 200

10x + 8y ≤ 240

Total profit, Z = 5x + 6y

The mathematical formulation of the given problem is

Maximize Z = 5x + 6y … (1)

subject to the constraints,

5x + 8y ≤ 200

5x + 4y ≤ 120

x, y ≥ 0 … (4)

Thus, e souvenirs Of type A and 20 souvenirs Of type B should be produced each day to get the
maximum pro t of Rs 160.

Page : 520 , Block Name : Exercise 12.2

Q8 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 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 pro t if he does not want to invest more than Rs 70 lakhs
and if his pro t on the desktop model is Rs 4500 and on portable model is Rs 5000.

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

x ≥ 0 and y ≥ 0

The cost of a desktop model is Rs 25000 and of a portable model is Rs 4000. However,

the merchant can invest a maximum of Rs 70 lakhs.

∴ 25000x + 40000y ≤ 7000000

5x + 8y ≤ 1400

∴ x + y ≤ 250

The monthly demand of computers will not exceed 250 units.

∴ 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

Thus, the mathematical formulation of the given problem is

Maximum Z = 4500x + 5000y

Thus, the mathematical formulation of the given problem is

Maximum Z = 4500x + 5000y

subject to the constraints,

5x + 8y ≤ 1400

x + y ≤ 250

x + y ≤ 250

x, y ≥ 0

The feasible region determined by the system of constraints is as follows.

The maximum value of Z Is 1150000 at (200, SO).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum
pro t of Rs 1150000.

Page : 520 , Block Name : Exercise 12.2

Q9 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 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 diet
that consists of mixture of these two foods and also meets the minimal nutritional requirements.

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 complied in a table as follows.

The cost of food F1 is Rs 4 per unit and of Food F2 is Rs 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 … (1)

subject to the constraints,

3x + 6y ≥ 80 … (2)

4x + 3y ≥ 100 … (3)

x, y ≥ 0 … (4)

The feasible region determined by the constraints is as follows.

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

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 Rs 104 .

Page : 520 , Block Name : Exercise 12.2

Q10 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
nds that she needs at least 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?

Let the farmer buy x kg of fertilizer F1 and y kg of fertilizer F2 . Therefore,

x ≥ 0 and y ≥ 0

The given information can be complied in a table as follows.

F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. However, the farmer

requires at least 14 kg of nitrogen.

∴ 10% of x + 5% of y ≥ 14
x y
+ ≥ 14
10 20

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.

6x 10y
+ ≥ 14
100 100

3x + 56y ≥ 700

Total cost of fertilizers, Z = 6x + 5y

The mathematical formulation of the given problem is

Minimize z = 6x + 5y … (1)

subject to the constraints,

Total cost of fertilizers, z = 6x + 5y

The mathematical formulation of the given problem is

Minimize Z = 6x + 5y … (1)

subject to the constraints,

2x + y ≥ 280 … (2)

3x + 5y ≥ 700 … (3)

x, y ≥ 0 … (4)

It can be seen that the feasible region is unbounded.

700
The corner points are ( , 0) , B(100, 80), and C(0, 280)
3

The values of z at these points are as follows.

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

Page : 520 , Block Name : Exercise 12.2

Q11 The corner points of the feasible region 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

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.

Page : 520 , Block Name : Exercise 12.2

Q1 Refer to Example 9. How many packets of each food should be used to maximise the amount of
vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

Let the diet contain x and y packets of foods P and Q respectively. Therefore

x ≥ 0 and y ≥ 0

The mathematical formulation of the given problem is as follows.

Maximize z = 6x + 3y … (1)

subject to the constraints,

4x + y ≥ 80

x + 5y ≥ 115

3x + 2y ≤ 150

x, y ≥ 0

The feasible region determined by the system of constraints is as follows.

The corner points of feasible region are A (15, 20), B (40, 15), and C (2, 72).
The values of at these corner points are as follows.

Thus, the maximum value Otz is 28S at (40, 15).

Therefore, to maximize the amount of vitamin A in the diet, 40 packets of food P and 15 packets of
food Q should be used. The maximum amount of vitamin A in the diet is 285 unit.

Page : 525 , Block Name : Miscellaneous Exercise

Q2 A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units
of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per
bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The
minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively.
Determine the number of bags of each brand which should be mixed in order to produce a mixture
having a minimum cost per bag? What is the minimum cost of the mixture per bag?

Answer. Let the farmer mix x bags Of brand p and v bags Of brand Q.
The given information can be compiled in a table as follows.

The given problem can be formulated as follows.

Minimize z = 250x + 200y … (1)

subject to the constraints,

3x + 1.5y ≥ 18

2.5x + 11.25y ≥ 45

2x + 3y ≥ 24

x, y ≥ 0

The feasible region determined by the system of constraints is as follows.

The corner points Of the feasible region are A (18, O), B (9, 2), C (3, 6), and D (O, 12).
The values ot z at these corner points are as follows.

As the feasible region is unbounded, therefore, 1950 may or may not be the minimum

value of z .

For this, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39, 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 5x + 4y < 39

Therefore, the minimum value of z is 2000 at (3, 6) .

Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to

minimize the cost to Rs 1950 .

Page : 526 , Block Name : Miscellaneous Exercise

Q3 A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture
contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin
contents of one kg food is given below:

One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which
will produce the required diet?

Let the mixture contain x kg of food x and y kg of food Y .

The mathematical formulation of the given problem is as follows.

Minimize z = 16x + 20y … (1)

subject to the constraints,

x + 2y ≥ 10

x + y ≥ 6

3x + y ≥ 8

x, y ≥ 0

The corner points of the feasible region are A (10, O), B (2, 4), C (I, 5), and D (O, 8).
The values of z at these corner points are as follows.

As the feasible region is unbounded, therefore, 112 may or may not be the minimum

value of z

For this, we draw a graph of the inequality, 16x + 20y < 112 or 4x + 5y < 28, 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 4x + 5y < 28

Therefore, the minimum value of z is 112 at (2, 4) .

Thus, the mixture should contain 2kg of food x and 4kg of food Y . The minimum cost of

the mixture is Rs 112 .

Page : 526 , Block Name : Miscellaneous Exercise

Q4 A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and
the time (in minutes) required for each toy on the machines is given below:

Each machine is available for a maximum of 6 hours per day. If the pro t on each toy of type A is Rs
7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be
manufactured in a day to get maximum pro t.

Let x and y toys of type A and type B respectively be manufactured in a day.

The given problem can be formulated as follows.

Maximize z = 7.5x + 5y … (1)

subject to the constraints,

2x + y ≤ 60

x ≤ 20

x ≤ 20

x, y ≥ 0

The corner points of the feasible region are A (20. 0). B (20. 20). C (15, 30). and 0 (0,
40).
The values of z at these corner points are as follows.

The maximum value of z s 262.5 at (15 , 30).
Thus, the manufacturer should manufacture 15 toys of type A and 30 toys Of type to maximize the
pro t.

Page : 526 , Block Name : Miscellaneous Exercise

Q5 An aeroplane can carry a maximum of 200 passengers. A pro t of Rs 1000 is made on each
executive class ticket and a pro t of Rs 600 is made on each economy class ticket. The airline reserves
at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by
economy class than by the executive class. Determine how many tickets of each type must be sold in
order to maximise the pro t for the airline. What is the maximum pro t?

Let the airline sell x tickets of executive class and y tickets of economy class.

The mathematical formulation of the given problem is as follows.

Maximize z = 1000x + 600y … (1)

subject to the constraints,

x + y ≤ 200

x ≥ 20

x ≥ 20

y − 4x ≥ 0

x, y ≥ 0

The corner points Of the feasible region are A (20, 80), B (40, 160), and c (20, 180).
The values of z at these corner points are as follows.

The maximum value Of z is 136000 at (40, 160).
Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to
maximize the pro t and the maximum pro t is Rs 136000.

Page : 526 , Block Name : Miscellaneous Exercise

Q6 Two godowns A and B have grain 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:

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

Let godown A supply x and y quintals of grain to the shops D and E respectively. Then,

(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) = (x + y − 60) quintals will be

transported from godown 8 to shop E and F respectively.

The given problem can be represented diagrammatically as follows.

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

The given problem can be formulated as

Minimize z = 2.5x + 1.5y + 410 … (1)

subject to the constraints,

x + y ≤ 100

x + y ≤ 100

x ≤ 60

y ≤ 50

y ≤ 50

x + y ≥ 0

The feasible region determined by the system of constraints is as follows.

The corner points are A (60, 0), B (60, 40), C (SO, 50), and D (10, SO).
The values of z at these corner points are as follows.

The minimum value of is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, SO quintals,
and 40 quintals respectively and from B to D. E, and F is 50 0 quintals, and 0

Page : 527 , Block Name : Miscellaneous Exercise

Q7 An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The
company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and
3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the

following table:

Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be
scheduled in order that the transportation cost is minimum? What is the minimum cost?

Answer. Let x and V litres Of Oil be supplied from A to the petrol pumps, D and E. Then, (7000 — x —
y) will supplied from A to petrol pump F.
The requirement at petrol pump D is 4500 L. Since x L are transported from depot A. the remaining
(4500 —x) will be transported from petrol pump B. Similarly. (3000 - V) L and 3500 - (7000 - x - y) - (x y
- 3500) L will be transported from depot B to petrol pump E and F respectively. The given problem can
be represented diagrammatically as follows.

x ≥ 0, y ≥ 0, and (7000 − x − y) ≥ 0

⇒ x ≥ 0, y ≥ 0, and x + y ≤ 7000

4500 − x ≥ 0, 3000 − y ≥ 0, and x + y − 3500 ≥ 0

⇒ x ≤ 4500, y ≤ 3000, and x + y ≥ 3500

cost of transporting 10 L of petrol = Re 1

Therefore, total transportation cost is given by,

7 6 3 3 4 2
z = × x + y + (7000 − x − y) + (4500 − x) + (3000 − y) + (x + y − 3500)
10 10 10 10 10 10

= 0.3x + 0.1y + 3950

The problem can be formulated as follows.

Minimize z = 0.3x + 0.1y + 3950 … (1)

subject to the constraints,

x + y ≤ 7000

x ≤ 4500

x ≤ 4500

y ≤ 3000

x + y ≥ 0

x, y ≥ 0

The corner points of the feasible region are A(3500, 0), B (4500, 0), C(4500, 2500), D

(4000, 3000), and E(500, 3000).

The values of z at these corner points are as follows.

The minimum value of is 4400 at (500, 3000).
Thus, the oil supplied from depot A is 500 L 3000 and 3500 L and from depot B is
4000 L, O L, and O L to petrol pumps D. E, and F respectively.
The minimum transportation cost iS RS 4400.

Page : 527 , Block Name : Miscellaneous Exercise

Q8 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in
kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table.
Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at
most 310 kg of chlorine.

If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each
brand should be used? What is the minimum amount of nitrogen added in the garden?

Let the fruit grower use x bags of brand P and y bags of brand Q.

The problem can be formulated as follows.

Minimize z = 3x + 3.5y … (1)

subiect to the constraints.

x + 2y ≥ 240

x + 2.5y ≥ 90

1.5x + 2y ≤ 310

x, y ≥ 0

The maximum value of z is 470 at (40, 100).

Thus, 40 bags of brand P and 100 bags of brand Q should be added to the garden to

minimize the amount of nitrogen.

The minimum amount of nitrogen added to the garden is 470kg.

Page : 527 , Block Name : Miscellaneous Exercise

Q9 Refer to Question 8. If the grower wants to maximise the amount of nitrogen added to the garden,
how many bags of each brand should be added? What is the maximum amount of nitrogen added?

Let the fruit grower use x bags of brand P and y bags of brand Q.

The problem can be formulated as follows.

Maximize z = 3x + 3.5y … (1)

subject to the constraints,

x + 2y ≥ 240 … (2)

x + 0.5y ≥ 90 … (3)

1.5x + 2y ≤ 310 … (4)

x, y ≥ 0 … (5)

The feasible region determined by the system of constraints is as follows.

The maximum value of z is 595 at (140, 50) .

Thus, 140 bags of brand P and 50 bags of brand Q should be used to maximize the

amount of nitrogen.

The maximum amount of nitrogen added to the garden is 595kg.

Page : 528 , Block Name : Miscellaneous Exercise

Q10 A toy company manufactures two types of dolls, A and B. Market research and available resources
have indicated that the combined production level should not exceed 1200 dolls per week and the
demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of
dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the
company makes pro t of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each
should be produced weekly in order to maximise the pro t?

Let x and y be the number of dolls of type A and B respectively that are produced per

week.

The given problem can be formulated as follows.

Maximize z = 12x + 16y … (1)

subject to the constraints,

x + y ≤ 1200 … (2)
x
y ≤ ⇒ x ≥ 2y … (3)
2

x − 3y ≤ 600 … (4)

x, y ≥ 0 … (5)

The maximum value of z is 16000 at (800, 400).

Thus, 800 and 400 dolls of type A and type B should be produced respectively to get the

maximum profit of Rs 16000.

Page : 528 , Block Name : Miscellaneous Exercise