Question

The maximum value of $z=6x+8y$ subject to $x-y\ge 0,x+3y\le 12,x\ge 0,y\ge 0$ is

• A. $72$
• B. $42$
• C. $96$
• D. $24$

Answer 1

Answer: (A)

We have, $z=6x+8y$
subject to constrants $x-y\ge 0,x+3y\le 12,x\ge 0$
$y\ge 0$
On taking given constraints as equations, we get the following graph

Intersecting point of the line $x-y=0$ and $x+8y=12$ is $B(3,3)$
Here, $OABO$ is the required reasible region
Whose corner points are $O(0,0),A(12,0)$ and $B(0,4)$ Now,

Comer points	$Z=6x+8y$
$O(0,0)$	$6\times 0+8\times 0=0$
$A(12,0)$	$6\times 12+8\times 0=72$ (maximum)
$B(3,3)$	$6\times 3+8\times 3=42$