# NCERT Solutions Mathematics Class 12 Chapter 12 Linear

## Preview text

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

Exercise 12.1

page no: 513

1. Maximise Z = 3x + 4y

Subject to the constraints:

Solution:

The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is given below

O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of

Z at these points are given below

Corner point

Z = 3x + 4y

O (0, 0)

0

A (4, 0)

12

B (0, 4)

16

Maximum

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

2. Minimise Z = −3x + 4y

subject to

.

Solution: The feasible region determined by the system of constraints,

is given below

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region

The values of Z at these corner points are given below

Corner point

Z = - 3x + 4y

O (0, 0)

0

A (4, 0)

-12

Minimum

B (2, 3)

6

C (0, 4)

16

Hence, the minimum value of Z is – 12 at the point (4, 0)

3. Maximise Z = 5x + 3y

subject to

.

Solution: The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤

10, x ≥ 0, and y ≥ 0, are given below

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

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

region. The values of Z at these corner points are given below

Corner point

Z = 5x + 3y

O (0, 0)

0

A (2, 0)

10

B (0, 3)

9

C (20 / 19, 45 / 19)

235 / 19

Maximum

Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19)

4. Minimise Z = 3x + 5y

such that

.

Solution: The feasible region determined by the system of constraints, x + 3y ≥ 3, x + y ≥ 2, and x, y ≥ 0 is given below

It can be seen that the feasible region is unbounded. The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2) The values of Z at these corner points are given below

Corner point

Z = 3x + 5y

A (3, 0)

9

B (3 / 2, 1 / 2)

7

Smallest

C (0, 2)

10

7 may or may not be the minimum value of Z because the feasible region is unbounded

For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check the resulting

half plane have common points with the feasible region or not

Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

Thus, the minimum value of Z is 7 at point B (3 / 2, 1 / 2)

5. Maximise Z = 3x + 2y

subject to

.

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

0, is given below

A (5, 0), B (4, 3), C (0, 5) and D (0, 0) are the corner points of the feasible region.

The values of Z at these corner points are given below

Corner point

Z = 3x + 2y

A (5, 0)

15

B (4, 3)

18

Maximum

C (0, 5)

10

Hence, the maximum value of Z is 18 at the point (4, 3)

6. Minimise Z = x + 2y

subject to

.

Solution:

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

is given below

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

A (6, 0) and B (0, 3) are the corner points of the feasible region

The values of Z at the corner points are given below

Corner point

Z = x + 2y

A (6, 0)

6

B (0, 3)

6

Here, the values of Z at points A and B are same. If we take any other point such as (2, 2)

on line x + 2y = 6, then Z = 6

Hence, 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

7. Minimise and Maximise Z = 5x + 10y

subject to

.

Solution: The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x ≥ 0, and y ≥ 0, is given below

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

A (60, 0), B (120, 0), C (60, 30), and D (40, 20) are the corner points of the feasible

region. The values of Z at these corner points are given

Corner point

Z = 5x + 10y

A (60, 0)

300

Minimum

B (120, 0)

600

Maximum

C (60, 30)

600

Maximum

D (40, 20)

400

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)

8. Minimise and Maximise Z = x + 2y

subject to

.

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

200, x ≥ 0, and y ≥ 0, is given below

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

A (0, 50), B (20, 40), C (50, 100) and D (0, 200) are the corner points of the feasible

region. The values of Z at these corner points are given below

Corner point

Z = x + 2y

A (0, 50)

100

Minimum

B (20, 40)

100

Minimum

C (50, 100)

250

D (0, 200)

400

Maximum

The maximum value of Z is 400 at point (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)

9. Maximise Z = − x + 2y, subject to the constraints:
. Solution: The feasible region determined by the constraints, below

is given

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

Here, it can be seen that the feasible region is unbounded.

The values of Z at corner points A (6, 0), B (4, 1) and C (3, 2) are given below

Corner point

Z = - x + 2y

A (6, 0)

Z = - 6

B (4, 1)

Z = - 2

C (3, 2)

Z = 1

Since the feasible region is unbounded, hence, z = 1 may or may not be the maximum

value

For this purpose, we graph the inequality, - x + 2y > 1, and check whether the resulting

half plane has points in common with the feasible region or not.

Here, the resulting feasible region has points in common with the feasible region

Hence, z = 1 is not the maximum value.

Z has no maximum value.

10. Maximise Z = x + y, subject to Solution:
The region determined by the constraints,

. is given below

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming
There is no feasible region and therefore, z has no maximum value.

NCERT Solutions Mathematics Class 12 Chapter 12 Linear Programming

Exercise 12.2

page no: 519

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

Vitamin A (units / Vitamin B (units /

Cost (Rs / kg)

kg)

kg

Food P

3

5

60

Food Q

4

2

80

Requirement (units /

8

11

kg)

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

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 