The minimum value of $\mathrm{Z}=5 x+8 y$ subject to $x+y \geq 5,0 \leq x \leq 4, y \geq 2, x \geq 0$ $y…
The minimum value of $\mathrm{Z}=5 x+8 y$ subject to $x+y \geq 5,0 \leq x \leq 4, y \geq 2, x \geq 0$ $y \geq 0$ is
40
36
31
20
Solution
Required area is shaded.
Co-ordinates of vertices are $C \equiv(4,1)$;
$D \equiv(4,2)$ and $P \equiv(3,2)$
$\mathrm{Z}=5 \mathrm{x}+8 \mathrm{y}$
$\therefore \quad Z_{(C)}=20+8=28$
$\mathrm{Z}_{(\mathrm{D})}=20+16=36$
$Z_{(P)}=15+16=31$
Minimum value will be 28 .
