The maximum value of $z=10 z+25 y$ subject to $0 \leq x \leq 3$, $0 \leq y \leq 3, x+y \leq 5$ occurs at the…
- $(3,2)$
- $(2,3)$
- $(4,3)$
- $(5,4)$
Solution
Required area is shaded.
Vertices of the required region are $0(0,0)$;
$\mathrm{A}(3,0) ; \mathrm{B}(3,2) ; \mathrm{C}(2,3) ; \mathrm{D}(0,3)$
We have to maximize objective function
$\begin{array}{lll}
\mathrm{Z}=10 \mathrm{x}+25 \mathrm{y} & \\
\therefore \quad & \mathrm{Z}_{(\mathrm{O})}=0+0 & =0 \\
\mathrm{Z}_{(\mathrm{A})} & =30+0 & =30 \\
\mathrm{z}_{(\mathrm{B})} & =30+50 & =80 \\
\mathrm{Z}_{(\mathrm{C})} & =20+75 & =95 \\
\mathrm{z}_{(\mathrm{D})} & =0+75 & =75
\end{array}$Asked in: MHT CET 2021 (22 Sep Shift 1)