Let the number of pizza slices be p, the number of hot dogs be q, and the number of barbeque sandwiches be r.
1) The problem is subject to the following constraints:
a) The cost of a slice of pizza = 6/8 = $0.75; the cost of a hot dog = $0.45; the cost of a barbeque sandwich = $0.9. The maximum amount Julia can spend for buying ingredients and preparing food = $1500. Therefore, the first constraint is:
b) The food items are sold twice during a match. This means the total available oven space in square inches = 2 x 16 x 4 x 3 x 144 = 55296 square inches. Now each pizza takes up 14 x 14 =196 square inches, therefore, each pizza slice takes up 196/8 = 24.5 square inches. Each hotdog takes up 16 square inches, and each sandwich takes up about 25 square inches. Therefore, the second constraint is:
c) Julia expects to sell at least as many slices of pizza as hot dogs and sandwiches put together. This gives the third constraint:
d) Further, she expects that the number of hot dogs sold will be twice the number of sandwiches. This means q≥2r
e) Finally, the number of any food item can never be negative. Therefore,
2) The aim here is to maximize Julia’s profit: z=Selling price-Cost price.
The selling price of a pizza slice = selling price of a hot dog = $1.5; the selling price of a sandwich = $2.25. Therefore, profits from the three items are $0.75, $1.05, and $1.35 respectively. Therefore, the overall profit function is z=0.75p+1.05q+1.35r which needs to be maximized.
3) Solving the LP model – Simplex Method:
Objective Function to be maximized: z=0.75p+1.05q+1.35r
Adding the appropriate slack variables a, b, c, and d, the initial table is then given by:
The pivot column is r. The corresponding ratios are 1666.67, 2211.84, 0, and 0 for the first four rows respectively. Now this gives the pivot element as 25. Continuing in this manner, the final optimal solution is found out to be: p = 1250; q = 1250; r = 0; a = 0; b = 5296; c = 0; and d = 1250. These values match with the Excel analysis.
This gives the maximum profit = $ 2250 (substituting appropriate values in objective function)
Now Julia pays $1000 as rent per day, and $100 as rent for the oven per match. Therefore the net profit she makes = 2250 – 1100 = $1150. Since this is > her goal of making $1000 a day, she should lease the booth.
If Julia were to borrow some additional money, then the first constraint 0.75p+0.45q+0.9r≤1500 would change. Now for each extra dollar she gets, the dual value is $1.50. However, she the borrowing has a limit since the oven can be used only twice and has a finite capacity. Considering this, the limit restriction given by the model = $1658.88. Therefore, the maximum money she can borrow = 1658.88 – 1500 = $158.88, which will yield an additional profit of (158.88 x 1.50) = $238.32.
Assuming she does borrow $158.88, giving away $100 dollars to get some extra help seems prudent. This will make things easier for her, while still satisfying her aim of earning at least $1000 at the end of the day.
The following are some reasonable uncertain factors to be considered:
- If the weather is not very good, the number of students turning up for a match may be drastically reduced. Even if they do attend the game, many might not want to walk to the food booth in unpleasant weather.
- There might be a problem with the oven.
In either case, Julia may be unable to save for the next day’s ingredients. So it is advisable to not use up the whole $1500; instead a carefully calculated amount can be set aside in case of emergency. This way, even if her profit reduces a little, she can ensure that she successfully runs her business for all the matches.