Solve the following linear programming problem graphically:

Maximize and minimize $Z = 60x + 15y$ subject to the constraints $x + y ≤ 50, 3x + y ≤ 90, x, y ≥ 0$.

Maximum $Z=1800$, Minimum $Z=0$

The correct answer is Option (2) → Maximum $Z=1800$, Minimum $Z=0$

We draw the lines $x + y = 50, 3x + y = 90$ and shade the region satisfied by the given inequalities. The shaded region in the adjoining figure shows the feasible region determined by the given constraints. We observe that the feasible region OABC is convex polygon and bounded. So we use the corner point method to calculate the maximum and minimum value of Z.

The coordinates of the corner points O, A, B, C are (0, 0), (30, 0), (20, 30) and (0,50) respectively. We evaluate $Z=60x + 15y$ at each of these points.

Hence, the minimum value of Z is 0 at the point (0, 0) and maximum value of Z is 1800 at the point (30, 0).