Question

Maximize $Z = 10 x_1 + 25 x_2$, 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 $Z = 10 x_1 + 25 x_2$
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 $O A B C D$ 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 $95$.