Question

Maximum and minimise Z =3x -4y subject to $x-2y\le 0,-3x+y\le 4,x-y\le 6$ and $x,y\ge 0$

Answer 1

Given LPP is,
maximum 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).
$\begin{array}{cc}\text{Corner points}& \text{Corresponding value of Z=3x-4y}\\ (0,0)& 0\\ (0,4)& -16\leftarrow Minimum\\ (12,6)& 12\leftarrow Maximum\end{array}$

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.