# 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 $O ( 0, 0 ) , A (4, 0 ) , \text{and }B (0, 4 )$. The values of $Z$ at these points are as follows.$\\$ $\begin{array}{|c|c|c|c|} \hline \text{Corner point} & Z=3x+4y & \\ \hline O(0,0)& 0 & \\ \hline A(4,0)& 12 &\\ \hline B(0,4) & 16 & \to \text{Maximum}\\ \hline \end{array}$$\\ Therefore, the maximum value of Z is 16 at the point B ( 0, 4 ) 2 Minimise Z =- 3x + 4y$$\\$ 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.$\\$

##### 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.$\\$ $\begin{array}{|c|c|c|c|} \hline \text{Corner point} & Z=3x+5y & \\ \hline A(3,0)& 9 &\\ \hline B(\dfrac{3}{2},\dfrac{1}{2})& 7& \to \text{Smallest}\\ \hline C(0,2) & 10 & \\ \hline \end{array}$$\\ \\ As the feasible region is unbounded, therefore, 7 may or may not be the minimum value of Z.\\ For this, we draw the graph of the inequality, 3x + 5y < 7 , and check whether the resulting half plane has points in common with the feasible region or not.\\ It can be seen that the feasible region has no common point with 3x + 5y < 7 Therefore, the minimum value of Z is 7 at (\dfrac{3}{2},\dfrac{1}{2}) 5 Maximise Z = 3x + 2y$$\\$ 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).$$\\ The values of Z at these corner points are as follows.\\ \begin{array}{|c|c|c|c|} \hline \text{Corner point} & Z=3x+2y & \\ \hline A(5,0)& 15 &\\ \hline B(4,3)& 18& \to \text{Maximum}\\ \hline C(0,5) & 10 & \\ \hline \end{array}$$\\$ $\\$ $\\$ Therefore, the maximum value of $Z$ is $18$ at the point $(4, 3).$

##### Solution :

The feasible region determined by the constraints,$\\$ $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x \geq 0,\text{and} y \geq 0$. is as follows.

The corner points of the feasible region are $\\$ $A ( 6, 0 ),$ and $B ( 20, 40 ).$$\\ . The values of Z at these corner points are as follows.\\ \begin{array}{|c|c|c|c|} \hline \text{Corner point} & Z=x+2y \\ \hline A(6,0) & 6 \\ \hline B(0, 3) & 6 \\ \hline \end{array}$$\\$ 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 $, x + 2y = 6$ $\\$

9   Maximise $Z =- x + 2y$ , subject to the constraints: $x \geq 3, x + y \geq 5, x + 2 y \geq 6, y \geq 0 .$

##### Solution :

The feasible region determined by the constraints, $x \geq 3, x + y \geq 5, x + 2 y \geq 6, \text{and}y \geq 0 .$ is as follows

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 as follows.$\\$ $\begin{array}{|c|c|c|c|} \hline \text{Corner point} & Z=-x+2y \\ \hline A(6,0)& z=-x+2y \\ \hline B(4,1) & z=-2 \\ \hline C(3,2) & z=1\\ \hline \end{array}$$\\$ 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.

10   Maximise $Z = x +y$ , subject to $x - y \leq - 1, - x + y \geq 0, x , y \leq 0 .$

##### Solution :

The region determined by the constraints, $x - y \leq - 1, - x + y \geq 0, x , y \leq 0 .$ is as follows.

There is no feasible region and thus, $Z$ has no maximum value.