The minimum value of the objective function $Z = x + 2y$ of an L.P.P. subject to constraints $2x + y ≥ 3,\frac{x}{2} + 2y ≥6,x ≥ 0, y ≥ 0$ is

6

The correct answer is Option (2) → 6

Given constraints

$2x+y\ge3$

$\frac{x}{2}+2y\ge6$

$x\ge0,\;y\ge0$

Find corner point by solving

$2x+y=3$

$\frac{x}{2}+2y=6$

From first $y=3-2x$

Substitute in second

$\frac{x}{2}+2(3-2x)=6$

$\frac{x}{2}+6-4x=6$

$\frac{x}{2}-4x=0$

$-\frac{7x}{2}=0$

$x=0$

$y=3$

Other feasible corner points on axes

On $x=0$, $\frac{x}{2}+2y\ge6\Rightarrow y\ge3$ gives $(0,3)$

On $y=0$, $\frac{x}{2}+2y\ge6\Rightarrow x\ge12$ gives $(12,0)$

Evaluate $Z=x+2y$

$Z(0,3)=6$

$Z(12,0)=12$

Minimum value of $Z$ is $6$ at $(0,3)$.