Linear Programming — Class 12 Maths Solution

Given LPP is, maximise and minimise $Z = 3x - 4y$

subject to $x - 2y \le 0,$ $- 3x + y \le 4,$ $x - y \le 6,$ $x,$ $y \ge 0$.

[on solving $x - y = 6$ and $x - 2y = 0$,

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).

For given unbounded region the minimum value of Z may or may not be -16. So,

for deciding this, we graph the inequality.

$3x - 4y < - 16$

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 $3x - 4y > 12$

and see that resulting open half plane has no common points with the feasible region

and hence maximum value 12 exist for $Z = 3x - 4y$.

Objective Questions