A) Step1: In this case Julia has three decisions variable: X1 numbers of pizza slices, X2 numbers of hot dogs, and X2 numbers of BBQ sandwiches. As in any linear programming (LP) problem, the first and most important step is that of problem formulation. Julia must sell to maximize her profit and help her decide if she should lease the booth or not. Therefore, I am going to estimate the profit for each item as difference between total revenue and total cost of the item.

For the first game, we have the following model:

Step 2: Objective Function

Maximize Z (Profit) = $0.75X1 + $1.05X2 +$1.35X3

Step 3:Constraints:

Budget = 0.75X1 + 0.45X2 + 0.90X3 ≤ $1,500  

Oven space = 24.5X1 + 16X2 + 25X3 ≤ 55,296 in²

Pizza sales =X1 ≥ X2 + X3   or X1 - X2 - X3 ≥ 0

                        Hot Dog sales vs. BBQ sandwiches = X2/X3 ≥ 2.0 or X2 - 2X3 ≥ 0

                                                     X1, X2, X3 ≥ 0

Money:

From excel, we get the optimum value as follows:

X1 = 1250; X2 = 1250 and X3 = 0 and Maximum value of Z = $2250

Julia should stock 1250 slices of pizza and 1250 numbers of Hot dogs. She need not to stock sandwiches.

Maximum Profit that can be expected is $2250.

Lease cost for the booth per game = $1000

Lease cost for the oven per game = $100

Net profit after all the expenses = 2250 - 1100 = $1150

Now it is clear that as per the strategy it is worth leasing the booth.

Oven space:

Space available = 3 x 4 x 16 = 192 sq. feet = 192 x 12 x 12 =27648 sq. inches

The oven will be refilled during half time.

Thus total space available = 27648 x 2 = 55296

      Space required for pizza slice to keep warm = 14 x 14/8 = 24.5 sq. inches

     Total Space required: 24.5 X1 + 16 X2 + 25 X3

     Constraint: 24.5 X1 + 16 X2 + 25 X3 ≤ 55296

     By using excel solver, this problem can be easily solved and we get:

                                                X1 = 1,250 slices of pizza

                                                X2 = 1,250 hot dogs

                                                X3 = 0 BBQ sandwiches

                                                Profit Z = $2,250

B)From excel solver, we can see that the dual value is $1.50 for each additional dollar and the allowable increase is 138.4. This means that each dollar added to the budget can increase a profit of $1.5 and the maximum allowable increase is 138.4. Therefore, Julia would increase her profit, if she borrows some money. So maximum amount that Julia can borrow from her friend to make profit is $138.4 and it will make an additional profit of 138.4 x 1.5 = $ 207.6.

It is the space factor that constraints her from borrowing even more money.

C) Julia should hire a friend to help her for $100 per game, which will help her to prepare all the food in such a short time. Her net profit per game will be 1150-100 = $1050. Still, it is worth

leasing the booth.

D)The biggest problem here is uncertain demand. She thinks that she will sell almost everything she can stock. If this is violated, then every thing will collapse.  According the result in (A), the net profit she can earn is $1150. It is above her target $1000. Therefore, she can suffer a profit of $150/- within her target. Now, as per the result in (C), it is physically difficult for her to prepare all the 1250 hot dogs and consequently she has to hire a friend for $100 per game. So, it is better to reduce the number of hotdogs to be prepared so that she can avoid hiring of the friend and at the same time reduce the risk of uncertain demand. If she reduces the number of hot dogs to 1108, it will reduce the profit only by $149.1. Still she can maintain her target of $1000 profit per game.