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
Answer & explanation
Correct answer: option 2
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 $(0,3)$.