Question

$Z=6x+21y$, subject to $x+2y\ge 3$, $x+4y\ge 4$, $3x+y\ge 3$, $x\ge 0$, $y\ge 0$. The minimum value of $Z$ occurs at

• A. $(4,0)$
• B. $(28,8)$
• C. $\left(2,\frac{1}{2}\right)$
• D. $(0,3)$

Answer 1

Answer: (C)

We have, minimize $Z=6x+21y$
Subject to $x+2y\ge 3,x+4y\ge 4,3x+y\ge 3,x\ge 0,y\ge 0$
Let $l_1:x+2y=3,l_2:x+4y=4,l_3:3x+y=3$
$l_4:x-0$ and $l_5:y=0$

For B : Solving $l_1$ and $l_2$, we get $B(2,1/2)$
For C : Solving $l_1$ and $l_3$, we get $C(0.6,1.2$)
Shaded portion is the feasible region, where $A(4,0)$,
$B\left(2,\frac{1}{2}\right)C(0.6,1.2),D(0,3)$.
Now, minimize $Z=6x+21y$
$Z$ at $A(4,0)=6(4)+21(0)=24$
$Z$ at $B\left(2,\frac{1}{2}\right)=6(2)+21(\frac{1}{2})=22.5$
$Z$ at $C(0.6,1.2)=6(0.6)+21(1.2)=3.6+25.2=28.8$
$Z$ at $D(0,3)=6(0)+21(3)=63$
Thus, $Z$ is minimized at $\left(2,\frac{1}{2}\right)$ and its minimum value is $22.5$.