The maximum value of $z=2 x+3 y$ subject to constraints $3 x-3 y ≥ 0,2 x+2 y ≤ 12, x ≥ 0, y ≥ 0$ occurs at the point
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) - $(3,3)$
$z=2 x+3 y$
$3 x-3 y ≥ 0,2 x+2 y ≤ 12$
$⇒x-y≥ 0,⇒x+y≤6$ $x, y ≥ 0$
finding intersection point
$x=y$
$⇒x+y=6$
$⇒x=y=3$
| corner points | $z=2 x+3 y$ |
| $A(0,0)$ | $Z_A=0$ |
| $B(3,3)$ | $Z_B=15$ |
| $C(6,0)$ | $Z_C=12$ |
maximum occurs at $B(3,3)$