Maximise and minimise $Z=3 x-4 y$ subject to $x-2 y \leq 0,-3 x+y \leq 4$, $x-y \leq 6$ and $x, y \geq 0$.
Given LPP is,
maximise and minimise $Z=3 x-4 y$ subject to $x-2 y \leq 0,-3 x+y \leq 4, x-y \leq 6, x, y \geq 0$.
[on solving $x-y=6$ and $x-2 y=0$, we get $x=12, y=6$ ]
From the shown graph, for the feasible region, we see that it is unbounded and coordinates of corner points are $(0,0),(12,6)$ and $(0,4)$.
| Corner points | Corresponding value of $Z=3x-4y$ |
|---|---|
| (0, 0) | 0 |
| (0, 4) | $-$16 $\leftarrow$ Minimum |
| (12, 6) | 12 $\leftarrow$ Maximum |
For given unbounded region the minimum value of $Z$ may or may not be $-16$ . So, for deciding this, we graph the inequality.
$$3 x-4 y<-16$$
and check whether the resulting open half plane has common points with feasible region or not.
Thus, from the figure it shows it has common points with feasible region, so it does not have any minimum value.
Also, similarly for maximum value, we graph the inequality $3 x-4 y>12$ and see that resulting open half plane has no common points with the feasible region and hence maximum value 12 exist for $Z=3 x-4 y$.
The corner points of the feasible region determined by the system of linear constraints are $(0,0),(0,40),(20,40),(60,20),(60,0)$. The objective function is $Z=4 x+3 y$. Compare the quantity in column $A$ and column $B$.
| Column A | Column B |
|---|---|
| Maximum of Z | 325 |
The feasible solution for a LPP is shown in following figure. Let $Z=3 x-4 y$ be the objective function. Minimum of $Z$ occurs at
Refer to question 27. Maximum of $Z$ occurs at
Refer to question 7, maximum value of $Z+$ minimum value of $Z$ is equal to