Point of intersection of $x=4$ and $3 x+2 y=18$ is $Q \equiv(4,3)$ Point of intersection of $\mathrm{y}=6$ and $3 \mathrm{x}+2 \mathrm{y}=18$ is $\mathrm{P} \equiv(2,6)$ Point $\mathrm{D} \equiv(4,0)$ and $\mathrm{C} \equiv(0,6)$ are as shown. The feasible region of th given L.P.P. is shaded portion CPQ D O. We have to maximize $Z=3 x+5 y$ Now, $\mathrm{Z}$ at $\mathrm{C}(0,6)=3(0)+5(6)=30$
$\mathrm{Z}$ at $\mathrm{P}(2,6) \quad=3(2)+5(6)=36$
$\mathrm{Z}$ at $\mathrm{Q}(4,3)=3(4)+5(3)=27$
$Z$ at $D(4,0)=3(4)+5(0)=12$
$\mathrm{Z}$ at $\mathrm{O}(0,0)=3(0)+5(0)=0$
Clearly the maximum value of $\mathrm{Z}$ is 36 at $\mathrm{P}(2,6)$