Julia's Food Booth

A) P: the #of pizza slices H: # of hot dogs B : # of barbeque sandwiches

Total revenue Total cost Profit
Pizza slice 1.50 6/8 slices = $0.75 0.75
Hot dog 1.50 0.45 1.05
Barbeque sandwich 2.25 0.90 1.35

Profit =Sell- Cost

The profit function: maximize Z = $0.75* x1 + $1.05 * x2 + $1.35 * x3
Maximize Z= 0 .75 p + 1.05H + 1.35B (the objective function)
Constraint 1 .75p + .45H + .9B <= 1500 ( cost) the area of one shelve in the warming oven is = (3*12)*(4*12)*16= 27648 sq. inches The oven will be refilled before half time- 27648 * 2 = 55296 .
Space required for pizza = 14 *14 = 196 sq. inches.
Space required for pizza slice = 196/ 8 = 24.50 sq. inches

Space for a hotdog=16 sq. inches

Space for a barbecue sandwich = 25 sq. inches

Constraint 2 24.5P + 16H + 25B <= 55296 ( space)
Constraint 3 p - H - B >= 0 selling at least as many slice of pizza as hot dogs and barbecue . . Together
Constraint 4 H - 2B >= 0 selling at least twice as many hot dogs as barbecue . Sandwiches p,H,B >= 0 . si>= 0
Based on QM for windows l solution the optimum solution:

Pizza (p) = 1250; Hotdogs(H) = 1250 and Barbecue sandwiches (B) = 0

Maximum value of Z = $2250

Julia should stock 1250 slices of pizza, 1250 hot dogs and no barbecue sandwiches.

Maximum Profit = $2250.

|Maximum Profit