Answer:
x₁ = 400 x₂ = 1000
z(max) = 380 $
Step-by-step explanation:
Production:
Flour yeast Almond paste Profit $
Bear claws B (x₁) 6 ou 1 ou 2 TS 0.2
Almond-filled ( x₂) 3 ou 1 ou 4 TS 0.3
Availability 6600 1400 4800
The Model:
z = 0.2*x₁ + 0.3*x₂ to maximize
Subject to
1.-Quantity of flour 6600 ou
6*x₁ + 3*x₂ ≤ 6600
2.-Quantity of yeast 1400 ou
1*x₁ + 1*x₂ ≤ 1400
3.-Quantity of Almond paste 4800 TS
2*x₁ + 4*x₂ ≤ 4800
General constraints:
x₁ ≤ 0 x₂ ≤ 0 both integers
After 6 iterations using an on-line solver. Optimal solution is:
x₁ = 400 x₂ = 1000
z(max) = 380 $