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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. $4x=3⇒x=\frac{3}{4}$ $y=\frac{3}{2}$
Max. value = $\frac{9}{4}$ |
