Julia's Food Booth

Decision Variables this problem includes three decision variables, representing the number of each food item to purchase: x1 = number of pizza slices to purchase x2 = number of hot dogs to purchase x3 = number of barbeque sandwiches to purchase

Objective Function The objective of Julia’s food booth is to maximize the total profit.

The total profit is the sum of the individual profits gained from pizza slices, hot dogs and barbeque sandwiches.

Profit derived from a pizza slice = $1.50 ($6/8 slices) = $1.50 - $0.75 = $0.75

Profit derived from a hot dog = $1.50 - $0.45 = $1.05

Profit derived from a barbeque sandwich = $$2.25 - $0.90 = $1.35

Thus, total profit, which we will define symbolically as Z, can be expressed mathematically as $0.75x1 + $1.05x2 + $1.35x3.

The objective function can be written as: maximize Z = $0.75x1 + $1.05x2 + $1.35x3.

Where Z = total profit for the first home game $0.75 x1 = profit from pizza slices

$1.05x2 = profit from hot dogs

$1.35x3 = profit from barbeque sandwiches

Model Constraints

The first constraint is budget constraint. Money available for purchase is $1,500. Cost for the purchase of per pizza slice is $0.75 (=$6/8 slices). Cost for the purchase of per hot dog is $0.45. Cost for the purchase of per barbeque sandwich is $0.90. Total cost for the purchase pizzas, hot dogs, and barbeque sandwiches is $0.75x1 + $0.45x2 + $0.90x3. Therefore, the budget constraint is expressed as: $0.75x1 + $0.45x2 + $0.90x3 = 2. These two constraints can be rewritten as: x1 - x2 - x3 >= 0 x2 - 2 x3 >= 0 Model Summary

The linear programming model for Julia’s food booth can be summarized as follows: maximize Z = 0.75x1 + 1.05x2 + 1.35x3 subject to: 0.75x1 + .0.45x2 + 0.90x3 = 0 x1, x2, x3 >= 0 (Non negativity constraint) Computer Solution with Excel can be used to solve linear programming problems. The decision variables are located in cells B12:B14. The profit is computed in cell B15 by the formula for profit, =...