Consider $Z=x+y $ subject to the constraints $2x+y≤3, 2x-y ≥0, x ≥ 0, y ≥0$ the maximum value of Z is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{9}{4}$
$2x-y=0$
$2x+y=3$
Adding both eq.
$4x=3⇒x=\frac{3}{4}$
$y=\frac{3}{2}$
| Corner points | (Z(x,y)= x + y) |
| (0, 0) | Z(0, 0) = 0 |
| (3/2, 0) | Z(3/2, 0) = 3/2 |
| (3/4, 3/2) | Z(3/4, 3/2) = 9/4 |
Max. value = $\frac{9}{4}$