Linear Programming

Class 12 NCERT

NCERT

1   Maximise $Z = 3x + 4y$ Subject to the constraints: $x + y \leq 4, x \geq 0, y \geq 0$

Solution :

The feasible region determined by the constraints, $x + y \leq 4, x \geq 0, y \geq 0 $, is as follows.$\\$ The corner points of the feasible region are $O ( 0, 0 ), A ( 4, 0 )$ , and $B ( 0, 4 )$ .$\\$ The values of $Z$ at these points are as follows.$\\$ table $\\$ Therefore, the maximum value of $Z$ is $16$ at the point $B ( 0, 4 )$

2   Minimise $Z =- 3x + 4y$$\\$ Subject to $ x + 2 y \leq 8,3 x + 2 y \leq 12, x \geq 0, y \geq 0.$

Solution :

The feasible region determined by the system of constraints, $x +2y \leq 8,3x + 2y \geq 12, x \leq 0$ and $y \leq 0,$ is as follows.$\\$ The corner points of the feasible region are $O ( 0, 0 ) , A ( 4, 0 ) , B (2, 3 ) ,$ and $C ( 0, 4 )$ $\\$ The values of $Z$ at these corner points are as follows.$\\$ table $\\$ Therefore, the minimum value of $Z$ is $-12$ at the point $( 4, 0 ).$

3   Maximise $Z = 5x + 3y$ $\\$ Subject to $3 x + 5 y \leq 15,5 x + 2 y \leq 10, x \geq 0, y \geq 0.$

Solution :

The feasible region determined by the system of constraints, $3x + 5y \leq 15, 5x + 2y \leq 10, x \geq 0$ and $y \leq 0 $, are as follows.$\\$ The corner points of the feasible region are $O ( 0, 0 ) , A (2, 0 ) , B ( 0, 3 )$ , and $C(\dfrac{20}{19},\dfrac{45}{19})$$\\$ The values of $Z$ at these corner points are as follows.$\\$ table $\\$ Therefore, the maximum value of $Z$ is $(\dfrac{235}{19})$ at the point $(\dfrac{20}{19},\dfrac{45}{19})$

4   Minimise $Z = 3x + 5y$$\\$ Such that $x + 3 y \leq 3, x + y \geq 2, x , y \geq 0$

Solution :

The feasible region determined by the system of constraints, $x + 3 y \leq 3, x + y \geq 2 $, and $x, y \geq 0 $ , is as follows.$\\$ It can be seen that the feasible region is unbounded.$\\$ The corner points of the feasible region are $A(3,0),B(\dfrac{3}{2},\dfrac{1}{2}),$ and $ C(0,2) $ $\\$ The values of $Z$ at these corner points are as follows.$\\$ table $\\$ As the feasible region is unbounded, therefore, $7$ may or may not be the minimum value of $Z$.$\\$ 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 of $Z$ is $7$ at $(\dfrac{3}{2},\dfrac{1}{2})$

5   Maximise $Z = 3x + 2y$$\\$ subject to $x + 2 y \leq 10,3 x + y \leq 15, x , y \geq 0 .$

Solution :

The feasible region determined by the constraints, $x + 2y \leq 10,3x + y \leq 15, x \geq 0$ and $y \geq 0$ , is as follows.$\\$ The corner points of the feasible region are $A (5, 0), B (4, 3), $and $C (0, 5).$$\\$ The values of $Z$ at these corner points are as follows.$\\$ table $\\$ Therefore, the maximum value of $Z$ is $18$ at the point $(4, 3).$