The maximum value of $\mathrm{z}=4 x+2 y$, subject to the constraints $3 x+4 y \geqslant 12, x+y \leqslant 5, x, y \geqslant 0$ is

The feasible region. lies on the origin side of $x+y=5$ and non-origin side of $3 x+4 y=12$, in the first quadrant.
The corner points of the feasible region are $\mathrm{A}(4,0), \mathrm{B}(5,0), \mathrm{C}(0,5)$ and $\mathrm{D}(0,3)$. $z=4 x+2 y$
At $\mathrm{A}(4,0), \mathrm{z}=16$ At B(5, 0), $z=20$ At C(0, 5), z=10 At $D(0,3), z=6$ $\therefore \quad$ Maximum value of $z$ is 20.