# NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1

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## NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1

Solve the following Linear Programming Problems graphically:

Ex 12.1 Class 12 Maths Question 1.
Maximize Z = 3x + 4y
subject to the constraints:
x + y ≤ 4,x ≥ 0,y ≥ 0.
Solution:
As 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 => $frac { x }{ 4 } +frac { y }{ 4 } =1$
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.

Ex 12.1 Class 12 Maths Question 2.
Minimize Z = -3x+4y
subject to x + 2y ≤ 8,3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Solution:
Objective function 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.
=> region x + 2y ≤ 8 lies on and below AB.

Ex 12.1 Class 12 Maths 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 constraints
are 3x + 5y≤15, 5x + 2y≤10,x≥0,y≥0

Ex 12.1 Class 12 Maths 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:

Ex 12.1 Class 12 Maths 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
=> $frac { x }{ 10 } +frac { y }{ 5 } =1$
Now (0,0) satisfies the inequation, therefore the half plane containing (0,0) is the required plane.
Again 3x+2y ≤ 15
Let 3x + y = 15
=> $frac { x }{ 5 } +frac { y }{ 15 } =1$
It is also satisfies by (0,0) and its required half plane contains (0,0).

Ex 12.1 Class 12 Maths 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
=> $frac { x }{ 6 } +frac { y }{ 3 } =1$
Here also (0,0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its comers 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.

Ex 12.1 Class 12 Maths Question 7.
Minimise and Maximise 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 constraints are x + 2y≤120,x+y≥60, x-2y≥0, x,y≥0

Ex 12.1 Class 12 Maths 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
=> $frac { x }{ 100 } +frac { y }{ 50 } =1$
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

Ex 12.1 Class 12 Maths 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

Ex 12.1 Class 12 Maths Question 10.
Maximize Z = x + y subject to x – y≤ -1, -x + y≤0,x,y≥0
Solution:
Objective function Z = x + y, constraints x – y≤ -1, -x + y≤0,x,y≥0

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