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

Max: 600, Min: 300

The correct answer is Option (2) → Max: 600, Min: 300

We find the intercepts for each constraint to plot the lines:

• Line 1 ($x + 2y = 120$):
• If $x = 0$, $y = 60$. Point: (0, 60)
• If $y = 0$, $x = 120$. Point: (120, 0)
• Line 2 ($x + y = 60$):
• If $x = 0$, $y = 60$. Point: (0, 60)
• If $y = 0$, $x = 60$. Point: (60, 0)
• Line 3 ($x - 2y = 0$ or $x = 2y$):
• Passes through the origin (0, 0).
• If $y = 30$, $x = 60$. Point: (60, 30).
• Non-negativity: $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

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.