The maximum value of `z=6x+8y` subject to `x-yge0,x+3yle12,xge0,yge0` is . . .

Options

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

Answer

A

Explanation

We have, z=6x+8y
subject to constrants x-y`ge0,x+3yle12,xge0,yge0`. On taking given constraints as equations,
we get the following graph

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

`therefore`Maximum value of Z is 72.