Answer:
z (max) = 6.15
x₁ = 1 x₂ = 3 x₃ = 6
Amount of beef leftover 2 lb
Amount of onions leftover 0
Amount of Special Sauce leftover 61
Amount of Hot Sauce leftover 9
Step-by-step explanation:
Ingredients Beef Onions Special S Hot S Profit
Wimpy (x₁) 1 2 5 0 0.6
Dial 911 (x₂) 1 2 2 5 0.55
Fire Bowl (x₃) 1.5 2 3 6 0.65
Available 15 20 90 60
Objective Function z:
z = 0.6*x₁ + 0.55*x₂ + 0.65*x₃ to maximize
Subject to:
1) Quantity of beef : 15
x₁ + x₂ + 1.5*x₃ ≤ 15
2) Quantity of onions: 20
2*x₁ + 2*x₂ + 2*x₃ ≤ 20
3) Quantity of Special sauce: 90
5*x₁ + 2*x₂ + 3*x₃ ≤ 90
4) Quantity of hot sauce: 60
0*x₁ + 5*x₂ + 6*x₃ ≤ 60
5) Condition: The number of servings for Fire Bowl must be at least 10% of the total number of servings for all three luncheon chili specials.
x₃ ≥ 0.1 ( x₁ + x₂ + x₃ ) or x₃ ≥ 0.1*x₁ + 0.1 *x₂ + 0.1*x₃
x₃ - 0.1*x₁ - 0.1 *x₂ - 0.1*x₃ ≥ 0
- 0.1*x₁ - 0.1 *x₂ + 0.9 *x₃ ≥ 0
6)Condition: The number of servings for Fire Bowl, however, cannot exceed the number of Dial 911 by more than 3.
x₃ - x₂ ≤ 3
7)the available number of servings for Dial 911 must be at least 2.
x₂ ≥ 2
General constraints:
x₁ ≥ 0 x₃ ≥ 0 all integers
With on-line solver solution is:
z (max) = 6.15
x₁ = 1 x₂ = 3 x₃ = 6
By sbstitution on the constraints
1) x₁ + x₂ + 1.5*x₃ ≤ 15 1 + 3 + 9 = 13
Amount of beef leftover 2 lb
2) 2*x₁ + 2*x₂ + 2*x₃ ≤ 20 2 + 6 + 12 = 20
Amount of onions leftover 0
3) 5*x₁ + 2*x₂ + 3*x₃ ≤ 90 5 + 6 + 18 = 29
Amount of Special Sauce leftover 61
4)0*x₁ + 5*x₂ + 6*x₃ ≤ 60 15 + 36 = 51
Amount of Hot Sauce leftover 9
