Chapter 12: Linear Programming
Q1. Maximize Z = 3x + 4y subject to: x + y ≤ 4, x ≥ 0, y ≥ 0.
Solution
Corner points of feasible region: (0,0), (4,0), (0,4).
At (0,0): Z = 0
At (4,0): Z = 12
At (0,4): Z = 16
Maximum Z = 16 at (0, 4)
Q2. Minimize Z = 5x + 3y subject to: 3x + 5y ≥ 15, 5x + 2y ≥ 10, x, y ≥ 0.
Solution
Corner points: (0, 5), (5/3, 2), (2, 0).
At (0,5): Z = 15
At (5/3, 2): Z = 25/3 + 6 = 43/3 ≈ 14.33
At (2,0): Z = 10
Minimum Z = 10 at (2, 0)
Q3. A manufacturer produces two products A and B. Product A requires 2 hours on machine I and 1 hour on machine II. Product B requires 1 hour on machine I and 3 hours on machine II. Machine I is available for 8 hours and machine II for 9 hours. If profit on A is Rs. 5 and on B is Rs. 3 per unit, find the maximum profit.
Solution
Let x = units of A, y = units of B.
Maximize Z = 5x + 3y
Subject to: 2x + y ≤ 8, x + 3y ≤ 9, x,y ≥ 0
Corner points: (0,0), (4,0), (3,2), (0,3)
At (0,0): Z = 0
At (4,0): Z = 20
At (3,2): Z = 15 + 6 = 21
At (0,3): Z = 9
Maximum Profit = Rs. 21 at (3, 2)
Q4. Maximize Z = x + 2y subject to: x + 2y ≤ 10, x + y ≤ 6, x, y ≥ 0.
Solution
Corner points: (0,0), (6,0), (2,4), (0,5).
At (0,0): Z = 0
At (6,0): Z = 6
At (2,4): Z = 2 + 8 = 10
At (0,5): Z = 10
Maximum Z = 10 at both (2,4) and (0,5)
Q5. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. F₁ costs Rs. 4/unit and contains 3 units of vitamin A and 5 units of minerals. F₂ costs Rs. 6/unit and contains 5 units of vitamin A and 4 units of minerals. Find the minimum cost.
Solution
Let x = units of F₁, y = units of F₂.
Minimize Z = 4x + 6y
Subject to: 3x + 5y ≥ 80, 5x + 4y ≥ 100, x,y ≥ 0
Corner points: (20, 0), (80/7, 100/7), (0, 25)
At (20,0): Z = 80
At (80/7, 100/7): Z = 320/7 + 600/7 = 920/7 ≈ 131.4
At (0,25): Z = 150
Minimum Cost = Rs. 80 at (20, 0)