The maximum value of $\mathrm{Z}=10 x+25 y$ subject to $0 \leq x \leq 3,0 \leq y \leq 3$, $x+y \leq 5, x…
The maximum value of $\mathrm{Z}=10 x+25 y$ subject to $0 \leq x \leq 3,0 \leq y \leq 3$,
$x+y \leq 5, x \geq 0, y \geq 0$ is
110
$100$
$120$
$95$
Solution
Given $\mathrm{Z}=10 \mathrm{x}+25 \mathrm{y}$ subject to $0 \leq \mathrm{x} \leq 3,0 \leq \mathrm{y} \leq 3, \mathrm{x}+\mathrm{y} \leq 5, \mathrm{x} \geq 0, \mathrm{y} \geq 0$
\begin{array}{|l|l|l|}
\hline Line & Point on \mathrm{X}-axis & Point on Y-axis \\
\hline \mathrm{x}+\mathrm{y}=5 & \mathrm{~A}(5,0) & \mathrm{B}(0,5) \\
\hline \mathrm{x}=3 & \mathrm{C}(3,0) & - \\
\hline \mathrm{y}=3 & - & \mathrm{D}(0,3) \\
\hline
\end{array}
Point of intersection of $x+y=5$ and $x=3$ is $F \equiv(3,2)$
Point of intersection of $x+y=5$ and $y=3$ is $G \equiv(2,3) .$
Feasible region OCFGDO is shaded.
We have $Z=10 x+25 y$
$\therefore Z_{(0)}=0+0=0$
$Z_{(C)}=30+0=30$
$Z_{(F)}=30+50=80$
$Z_{(G)}=20+75=95$
$Z_{(D)}=0+75=75$
