Linear Programming — Practice

Question 1

Maximise Z = 3x + 4y subject to: x + y ≤ 4; x + 2y ≤ 6; x ≥ 0; y ≥ 0.

Accepted Answer

CORNER POINTS: (1) (0,0): from x≥0, y≥0. (2) (4,0): from x+y=4 and y=0. (3) (0,3): from x+2y=6 and x=0 (y=3). (4) Intersection of x+y=4 and x+2y=6: subtract: y=2, x=2 → (2,2). FEASIBLE REGION: bounded polygon with vertices (0,0), (4,0), (2,2), (0,3). Z VALUES: Z(0,0)=0. Z(4,0)=12. Z(2,2)=6+8=14. Z(0,3)=12. MAXIMUM Z = 14 at (2, 2).

Question 2

A factory makes two products A and B. Product A requires 2 hours of machine time and 1 hour of labour; product B requires 1 hour of machine time and 2 hours of labour. Machine time available: 10 hours; labour available: 8 hours. Profit: A gives ₹4 per unit, B gives ₹3 per unit. How many units of A and B maximise profit?

Accepted Answer

Let x = units of A, y = units of B. OBJECTIVE: Maximise Z = 4x + 3y. CONSTRAINTS: Machine: 2x + y ≤ 10. Labour: x + 2y ≤ 8. Non-negativity: x ≥ 0, y ≥ 0. CORNER POINTS: (0,0), (5,0) [2x=10, y=0], (0,4) [x=0, 2y=8]. Intersection of 2x+y=10 and x+2y=8: multiply second by 2: 2x+4y=16. Subtract: 3y=6 → y=2, x=4. Corner: (4,2). Z VALUES: Z(0,0)=0. Z(5,0)=20. Z(4,2)=16+6=22. Z(0,4)=12. MAXIMUM Z = ₹22 at x=4, y=2. Produce 4 units of A and 2 units of B.

Question 3

Minimise Z = 5x + 10y subject to: x + 2y ≥ 120; x + y ≥ 60; x − 2y ≤ 0; x ≥ 0; y ≥ 0.

Accepted Answer

CONSTRAINTS (all ≥ → feasible region is ABOVE these lines — likely unbounded): (1) x+2y=120: intercepts (120,0) and (0,60). (2) x+y=60: intercepts (60,0) and (0,60). (3) x−2y=0 → x=2y. (4) x≥0, y≥0. CORNER POINTS: Find intersections in feasible region. Intersection of x+2y=120 and x−2y=0: add: 2x=120 → x=60, y=30. Point (60,30). Intersection of x+y=60 and x=2y: 2y+y=60 → y=20, x=40. Point (40,20). Intersection of x+2y=120 and x+y=60: subtract: y=60, x=0. Point (0,60). EVALUATE Z: Z(60,30) = 300+300=600. Z(40,20) = 200+200=400. Z(0,60) = 0+600=600. Minimum Z = 400 at (40,20). CHECK (unbounded): Open half-plane 5x+10y < 400 — test any interior point. At (50,10): 250+100=350 < 400, but check if (50,10) is in feasible region: x+2y=70 < 120 ✗ (not feasible). No feasible point gives Z < 400. Minimum Z = 400 at x=40, y=20. ✓

Practice — Linear Programming

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