Answer:
Z max = 1500
x₁ = 15
x₂ = 0
Explanation:
From problem statement:
Plant A Plant B Profit $/each
Product 1 (Doors ) x₁ 2 4 100
Product 2 (windows ) x₂ 3 4 80
Product 2 (windows ) x₂ above 10 is scrap
Objective function to maximize
z = 100*x₁ + 80*x₂
Subject to:
Plant A capacity
2*x₁ + 3*x₂ ≤ 40
Plant B capacity
4*x₁ + 4*x₂ ≤ 60
Windows condition:
0*x₁ + x₂ ≤ 10
Model:
Max z = 100*x₁ + 80*x₂
Subject to:
2*x₁ + 3*x₂ ≤ 40
4*x₁ + 4*x₂ ≤ 60
x₂ ≤ 10
x₁ , x₂ ≥ 0 and integers
Afther 6 iterations using Simplex Solver AtomZmath
Z max = 1500
x₁ = 15
x₂ = 0