The corner points of the feasible region of the LPP: Minimize z = -50…

The correct answer is Option (3) → $(0,3), (0,5), (1,0), (6,0)$

Constraints:

$2x-y\ge -5 \;\Longrightarrow\; y\le 2x+5$

$3x+y\ge 3 \;\Longrightarrow\; y\ge 3-3x$

$2x-3y\le 12 \;\Longrightarrow\; y\ge \frac{2x-12}{3}$

$x\ge 0,\; y\ge 0$

Find boundary intersections that lie in the feasible region.

On the line $y=0$:

$3-3x\le 0 \Longrightarrow x\ge 1$

$\frac{2x-12}{3}\le 0 \Longrightarrow x\le 6$

Hence two corner points: $(1,0)$ and $(6,0)$.

On the line $x=0$:

$y\le 5,\; y\ge 3,\; y\ge -4$

Hence two corner points: $(0,3)$ and $(0,5)$.

No other intersection of boundary lines lies in the first quadrant, so the feasible region has four corner points.

Final answer: the corner points are $(0,3),\,(0,5),\,(1,0),\,(6,0)$