BCS-012 — Group 4 Solutions (Question 6 — Linear Programming)

(d) Minimize Z = 4x + 3y, subject to:

2x+y ≥ 40, x+2y ≥ 50, x+y ≥ 35, x ≥ 0, y ≥ 0

Step 1: Find corner points of the feasible region

(intersection of boundary lines, checked for feasibility)

• 2x+y=40 and x=0 ⟹ (0, 40)
• 2x+y=40 and x+y=35 ⟹ subtract: x = 5, y = 30 ⟹ (5, 30)
• x+y=35 and x+2y=50 ⟹ subtract: y = 15, x = 20 ⟹ (20, 15)
• x+2y=50 and y=0 ⟹ (50, 0)

Each of these points is checked to satisfy all three inequalities (they do), so the feasible region (unbounded) has corner points:

(0,40), (5,30), (20,15), (50,0)

Step 2: Evaluate Z = 4x+3y at each corner point

Since the feasible region is unbounded, we additionally confirm that no point in the region gives a value of Z less than 110 (the half-plane 4x+3y < 110 does not intersect the feasible region).

Answer: Minimum value of Z = 110, attained at x = 5, y = 30.