NCERT Solutions for Maths Class 12 Exercise 12.1

# NCERT Solutions for Maths Class 12 Exercise 12.1

Hello Students. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 12.1.

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NCERT Solutions for Maths Class 12 Exercise 12.1

Solve the following Linear Programming Problems graphically:

Maths Class 12 Ex 12.1 Question 1.

Maximize Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

Solution:

Since x ≥ 0, y ≥ 0, therefore we shall shade the other inequalities in the first quadrant only. Now, consider x + y ≤ 4.
Let x + y = 4
x/4 + y/4 = 1
Thus, the line has 4 and 4 as intercepts along the axes. Now, (0, 0) satisfies the inequation, i.e., 0 + 0 ≤ 4. Now, shaded region OAB is the feasible solution.

Maths Class 12 Ex 12.1 Question 2.

Minimize Z = –3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

Solution:

The objective function is Z = –3x + 4y
Constraints are x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
(i) Consider the line x + 2y = 8. It pass through A (8, 0) and B (0, 4), putting x = 0, y = 0 in x + 2y ≤ 8, 0 ≤ 8, which is true.
the region x + 2y ≤ 8 lies on and below AB.

Maths Class 12 Ex 12.1 Question 3.

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

Solution:

The objective function is Z = 5x + 3y

The constraints are 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

Maths Class 12 Ex 12.1 Question 4.

Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

Solution:

For plotting the graph of x + 3y = 3, we have the following table:

Maths Class 12 Ex 12.1 Question 5.

Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

Solution:

Consider x + 2y ≤ 10
Let x + 2y = 10
x/10 + y/5 = 1
Now, (0, 0) satisfies the inequation, therefore, the half plane containing (0, 0) is the required plane.
Again, 3x + y ≤ 15
Let 3x + y = 15
x/5 + y/15 = 1
It is also satisfies by (0, 0) and its required half plane contains (0, 0).

Maths Class 12 Ex 12.1 Question 6.

Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

Solution:

Consider 2x + y ≥ 3
Let 2x + y = 3
y = 3 – 2x

(0, 0) is not contained in the required half plane as (0, 0) does not satisfy the inequation 2x + y ≥ 3.
Again, consider x + 2y ≥ 6
Let x + 2y = 6
x/6 + y/3 = 1
Here also, (0, 0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its corners are A (6, 0) and B (0, 3). At A, Z = 6 + 0 = 6
At B, Z = 0 + 2 × 3 = 6
We see that at both points, the value of Z = 6 which is minimum. In fact, at every point on the line AB makes Z = 6, which is also minimum.

Show that the minimum of Z occurs at more than two points:

Maths Class 12 Ex 12.1 Question 7.

Minimize and Maximize Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0

Solution:

The objective function is Z = 5x + 10y

The constraints are x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0

Maths Class 12 Ex 12.1 Question 8.

Minimize and maximize Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200, x, y ≥ 0.

Solution:

Consider x + 2y ≥ 100
Let x + 2y = 100
x/100 + y/50 = 1
Now, x + 2y ≥ 100 represents which does not include (0, 0) as it does not made it true.
Again, consider 2x – y ≤ 0
Let 2x – y = 0 or y = 2x

Maths Class 12 Ex 12.1 Question 9.

Maximize Z = –x + 2y subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

Solution:

The objective function is Z = –x + 2y.
The constraints are x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0

Maths Class 12 Ex 12.1 Question 10.

Maximize Z = x + y subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0

Solution:

The objective function is Z = x + y

The constraints are x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0