Question

$Z=4x_1+5x_2$, subject to $2x_1+x_2\ge 7$, $2x_1+3x_2\le 15$, $x_2\le 3$, $x_1$, $x_2\ge 0$. The minimum value of $Z$ occurs at

• A. $(3.5,0)$
• B. $(3,3)$
• C. $(7.5,0)$
• D. $(2,3)$

Answer 1

Answer: (A)

We have, minimize $Z=4x_1+5x_2$
subject to $2x_1+x_2\ge 7,2x_1+3x_2\le 15,x_2\le 3,x_1,x_2\ge 0$
Let $l_1$ : $2x_1+x_2=7$
$l_2:2x_1+3x_2=15$
$l_3:x_2=3,l_4:x_1=0$ and $l_5:x_2=0$

For C: Solving $l_2$ and $l_3$, we get $C(3,3)$
For D : Solving $l_1$ and $l_3$, we get $D(2,3)$
Shaded portion $ABCD$ is the feasible region where
$A(3.5,0),B(7.5,0),C(3,3),D(2,3)$
Now, minimize $Z=4x_1+5x_2$
$Z$ at $A(3.5,0)=4(3.5)+5(0)=14$
$Z$ at $B(7.50)=4(7.5)+5(0)=30$
$Z$ at $C(3,3)=4(3)+5(3)=27$
$Z$ at $D(2,3)=4(2)+5(3)=23$
Thus, $Z$ is minimized at $A(3.5,0)$ and its minimum value is $14$.