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Answer :
a. A linear programming model for this problem is 6x + y ≤ 1500, 5x + y ≥ 1600, 5x - 1.5y ≥ 0, x ≥ 0, y ≥ 0.
b. A graph of the feasible area with all the possible extreme points (1125/7, 3750/7), (250, 0), (48, 160), and (80, 0) is shown in the picture below.
c. The optimal solution point is (80, 0) and the optimal objective function value is $6.4.
Part a.
Based on the information provided above, the variables for the problem can be defined as follows;
Let the variable x represent the number of bottles of wine.
Let the variable y represent the number of bottles of beer.
Since a bottle of wine costs $6 and a bottle of beer costs $1, and the Robinsons have budgeted $1,500 for wine and beer, an inequality that defines this constraint is given by;
6x + y ≤ 1500
Since the Robinsons are planning on 400 guests at the reception, and they estimate that they need at least four servings, we have;
Total servings = 400 × 4 = 1600 servings.
5x + y ≥ 400
Also, they estimated that more than 50% guests will prefer wine to beer. Therefore, this constraint is given by;
5x ≥ (1 + 50/100)y
5x ≥ 1.5y
5x - 1.5y ≥ 0
Lastly, the caterer advised them that typically 8% of the wine and 15% of the beer will be left over. Therefore, the objective function for the minimization is given by;
Z = 0.08x + 0.15y
Non-negativity constraints;
x ≥ 0
y ≥ 0
Part b.
By critically observing the graph of the constraints above, the vertices include;
(1125/7, 3750/7), (250, 0), (48, 160), and (80, 0)
Part c.
Next, we would find the minimum cost as follows;
Z = 0.08(1125/7) + 0.15(3750/7) = $93.21.
Z = 0.08(250) + 0.15(0) = $20.
Z = 0.08(48) + 0.15(160) = $27.84.
Z = 0.08(80) + 0.15(0) = $6.4.
In conclusion, the optimal solution point is (80, 0) and the optimal objective function value is $6.4.
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