Question

$Z=8x+10y$, subject to $2x+y\ge 7$, $2x+3y\ge 15$, $y\ge 2$, $x\ge 0$, $y\ge 0$. The minimum value of $Z$ occurs at

• A. $(4.5,2)$
• B. $(1.5,4)$
• C. $(0,7)$
• D. $(7,0)$

Answer 1

Answer: (B)

We have, $Z=8x+10y$,
subject to $2x+y\ge 7,2x+3y\ge 15,y\ge 2,x\ge 0,y\ge 0$.
Let $l_1:2x+y=7,l_2:2x+3y=15,l_3:y=2$
$l_4:x=0,l_5:y=0$

For A: Solving $l_2$ and $l_3$, we get $A(4.5,2)$
For B : Solving $l_1$ and $l_2$, we get $B(1.5,4)$
Shaded portion is the feasible region, where
$A(4.5,2),B(1.5,4),C(0,7)$
Now, minimize $Z=8x+10y$
$Z$ at $A(4.5,2)=8(4.5)+10(2)=56$
$Z$ at $B(1.5,4)=8(1.5)+10(4)=52$
$Z$ at $C(0,7)=8(0)+10(7)=70$
Thus, $Z$ is minimized at $B(1.5,4)$ and its minimum value is $52$.