Z = $ .75(X1) + $1.05(X2) + $1.35(X3)

Given the following remains true:

$ .75(X1) + $1.05(X2) + $1.35(X3) =0 and,X2/X3 >=2; Solve for 0 for Excel: X2 >= 2(X3); X2 – 2(X3)>=0

Where X1, X2, and X3 are Pizza, Hotdog and BBQ Sandwiches respectively and are greater than 0. Based on the above LP model, Julie is expected to earn a profit of $2,250.00. After paying for rental lease, she has earned a net profit of $1250.00. The model suggests that she rents the booth and sell only pizza and Hotdog due to her spacing constraints. This will be the best option to achieve optimal results.

B. If Julia were to borrow more money to purchase more ingredients this would change her above profit. Any change in a coefficient in a parameter is carefully analyzed using a sensitivity analysis. This analysis identifies any effect an independent variable might have on Julia’s given constraints, in this case, her budget. This increase will generate an increase in product availability and also profitability.

Based on the sensitivity analysis, the upper limit of the sensitivity range for budget is $158.00 allowing her to have a maximum budget of $1658.00. Her shadow price reflects an increase of $1.50 for any additional resource. Therefore if she borrows the maximum of $158.88, she will yield an increase in profit of $238.32 and total profit of $2,488.32.

However, not all of her constraints are affected, as she still has to work within her specified area limits. This will pose a maximum of