$$
\text {Graph of given LPP is. }
$$


Solving $x-y=6$ and $x-2 y=0$,
we get $x=12, y=6$
The graph is unbounded and coordinates of corner points $\operatorname{are}(0,0),(12,6)$ and $(0,4)$
\begin{array}{|l|l|}
\hline Corner points & \mathbf{Z}=\mathbf{3} \mathbf{x}-\mathbf{4} \mathbf{y} \\
\hline(0,0) & 0 \\
(0,4) & -15 \leftarrow Minimum \\
(12,6) & 16 \leftarrow Maximum \\
\hline
\end{array}
Consider
$$
3 x-4 y < -16
$$
On checking we have that It has common points with feasible region, so it does not have any minimum value.
Similarly for maximum value, the inequality $3 x-4 y>12$
No common points with the feasible region and hence maximum value 12 exist for $\mathrm{Z}=3 \mathrm{x}-4 \mathrm{y}$.