Maximize and minimize $Z = 5x + 10y$ subject to the constraints $x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0$.

Minimum $Z=300$ at $(60,0)$; Maximum $Z=600$ at $(120,0)$

The correct answer is Option (2) → Minimum $Z=300$ at $(60,0)$; Maximum $Z=600$ at $(120,0)$

Maximize and minimize $Z = 5x + 10y$ subject to the constraints $x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0$.

Draw the lines $x + 2y 120, x + y = 60$, and $x - 2y = 0$; shade the region satisfied by the given inequalities. The feasible region is polygon ABCD, which is convex and bounded. Corner points of feasible region are $A(60, 0), B(120, 0), C(60, 30)$ and $D(40, 20)$.

The values of $Z = 5x + 10y$ at the points A, B, C and D are 300, 600, 600 and 400 respectively. Minimum value = 300 at A(60, 0), maximum value = 600 at B(120, 0) and C(60, 30). In fact, all points on the line segment BC give the same maximum value = 600.