Question

The maximum value of $z=9x+11y$ subject to $3x+2y\le 12,2x+3y\le 12,x\ge 0,y\ge 0$ is $\dots \dots$

• A. $44$
• B. $54$
• C. $36$
• D. $48$

Answer 1

Answer: (D)

We have, $Z=9x+11y$ Subject to $3x+2y\le 12,2x+3y\le 12$ On taking given constraints as equation, we get the following graph.

Intersection point of lines $3x+2y=12$ and $2x+3y=12$ is $C\left(\frac{12}{5},\frac{12}{5}\right)$
Here, $OACBO$ is the feasible region whose corner points are $O(0,0),A(4,0),C\left(\frac{12}{5},\frac{12}{5}\right)$ and $B(0,4)$.

Corner points	$Z=9x+11y$
$O(0,0)$	$0+0=0$
$A(4,0)$	$9\times 4+11\times 0=36$
$B(0,4)$	$9\times 0+11\times 4=44$
$C\left(\frac{12}{5},\frac{12}{5}\right)$	$9\times \frac{12}{5}+11\times \frac{12}{5}=48$ (maximum)