Respuesta :
Answer:
z(max) = 5000
x₁ = 2
x₂ = 32
Step-by-step explanation:
Time Assemble Finish Inspection Profit
Small x₁ 3 2 1 100
Large x₂ 4 3 1,5 150
Availablity 140 100 50
Objective function:
z = 100*x₁ + 150*x₂ to maximize
Subject to:
First constraint assembly capacity 140 hours
3*x₁ + 4*x₂ ≤ 140
Second constraint finishing capacity 100 hours
2*x₁ + 3*x₂ ≤ 100
Third constraint inspection capacity 50 hours
x₁ + 1,5*x₂ ≤ 50
Model:
z = 100*x₁ + 150*x₂ to maximize
Subject to:
3*x₁ + 4*x₂ ≤ 140
2*x₁ + 3*x₂ ≤ 100
x₁ + 1,5*x₂ ≤ 50
General constraints x₁ ≥ 0 x₂ ≥ 0 and integers
With the use of AtomZmath.com (online solver)
After 6 iterations the solution is:
z(max) = 5000
x₁ = 2
x₂ = 32
