Solve the following linear programming problem graphically: Maximize…

The correct answer is Option (3) → Max: 1800, Min: 0

Identify the Boundary Lines

First, we convert the inequalities into equations to find the intercepts for each boundary line:

• Constraint 1: $x + y = 50$
• If $x = 0$, $y = 50$. Point: (0, 50)
• If $y = 0$, $x = 50$. Point: (50, 0)
• Constraint 2: $3x + y = 90$
• If $x = 0$, $y = 90$. Point: (0, 90)
• If $y = 0$, $3x = 90 ⇒x = 30$. Point: (30, 0)
• Non-negativity: $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

Find the Intersection Point

We find the point where the two lines $x + y = 50$ and $3x + y = 90$ intersect:

• From the first equation: $y = 50 - x$
• Substitute into the second equation: $3x + (50 - x) = 90$
• $2x + 50 = 90 ⇒2x = 40 ⇒\mathbf{x = 20}$
• Substitute $x$ back: $y = 50 - 20 ⇒\mathbf{y = 30}$

The intersection point is (20, 30).

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.

Corner point	Value of objective function $Z=60x+15y$
(0, 0)	0 → Smallest
(30, 0)	1800 → Largest
(20, 30)	1650
(0, 50)	750

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).