We have, z = 6x +8y
Subject to constraints $x-y\ge 0,x+3y\le 12,x\ge 0,y\ge 0$ .
On taking given constraints as equations, we get the following graph

Intersecting point of the line $x-y=0$ and $x+3y=12$ is B (3, 3).
Here, OABO is the required feasible region
Whose corner points are O (0, 0), A (12, 0) and B (0, 4) Now,
Corner points $Z=6x+8y$
O (0, 0) $6\times 0+8\times 0=0$
A (12, 0) $6\times 12+8\times 0=72$
(maximum)
B (3, 3) $6\times 3+8\times 3=42$
$\therefore$ Maximum value of Z is 72.