For the linear programing problem,
Minimize $(Z) = 60x + 30y$
subject to: $2x - y ≥-5; 3x + y ≥ 3; 2x-3y≤12; x, y ≥0$
the optimal value of $Z$ is

60

The correct answer is Option (3) → 60 **

Minimize: $Z=60x+30y$

Subject to:

$2x-y\ge -5,\;3x+y\ge 3,\;2x-3y\le 12,\;x\ge0,\;y\ge0$

Feasible region: $y\le 2x+5$, $y\ge 3-3x$, $y\ge \frac{2}{3}x-4$, $x\ge0$, $y\ge0$.

Thus relevant vertices: $(0,3)$, $(0,5)$, $(1,0)$, $(6,0)$.

Evaluate $Z$ at corner points:

$Z(0,3)=60\cdot0+30\cdot3=90$

$Z(0,5)=60\cdot0+30\cdot5=150$

$Z(1,0)=60\cdot1+30\cdot0=60$

$Z(6,0)=60\cdot6+30\cdot0=360$

Minimum value is $Z=60$ at $(1,0)$.

Optimal value of $Z$ is $60$.