Consider $Z=x+y $ subject to the constraints $2x+y≤3, 2x-y ≥0, x ≥ 0, y ≥0$ the maximum value of Z is :

$\frac{3}{2}$ $\frac{5}{2}$ $\frac{7}{4}$ $\frac{9}{4}$

$\frac{9}{4}$

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 = $\frac{9}{4}$