Question

Maximize $Z=10x_1+25x_2$, subject to $0\le x_1\le 3$,
$0\le x_2\le 3,x_1+x_2\le 5$.

• A. $80$ at $(3,2)$
• B. $75$ at $(0,3)$
• C. $30$ at $(3,0)$
• D. $95$ at $(2,3)$

Answer 1

Answer: (D)

We have, maximize $Z=10x_1+25x_2$
Subject to $0\le x_1\le 3,0\le x_2\le 3,x_1+x_2\le 5$
Let $l_1:x_1=3$,
$l_2:x_2=3$
$l_3:x_1+x_2=5$,
$l_4:x_1=0$ and $l_5:x_2=0$

For B : Solving $l_1$ and $l_3$, we get $B(3,2)$
For C : Solving $l_2$ and $l_3$, we get $C(2,3)$
Shaded portion $OABCD$ is the feasible region,
where $O(0,0),A(3,0),B(3,2),C(2,3),D(0,3)$
Now maximize $Z=10x_1+25x_2$
$Z$ at $0(0,0)-10(0)+25(0)=0$
$Z$ at $A(3,0)=10(3)+25(0)=30$
$Z$ at $B(3,2)=10(3)+25(2)=80$
$Z$ at $C(2,3)=10(2)+25(3)=95$
$Z$ at $D(0,3)=10(0)+25(3)=75$
Thus, $Z$ is maximized at $C(2,3)$ and its maximum value is $95$.