Julias Food Booth

Complete the Julia’s Food Booth” Case Problem on page 109 of the text. Address each of the issues A – D according the instructions given.

A. Formulate and solve an L.P. model for this case:

Let X1= # of slices of pizza

X2= # of Hot Dogs

X3= # of Barbecue sandwiches

Max Z= 0.75X1 + 1.05X2 + 1.35X3

S.t. 0.75X1 + 0.45X2 + 0.90X3 ≤ 1,500 (Cost)

24.50X1 + 16X2 + 25X3 ≤ 55,296 (Oven space)

X1 – X2 – X3 ≥ 0 (Selling X1, X2, X3)

X2 - 2X3 ≥ 0 (Selling hot dogs & barbecue sandwich)

X1, X2, X3 ≥0

Constraints 1: Cost per Pizza slice= $6 for 8 slices= 6/8= 0.75

Cost for a Hot dog= 0.45

Cost for Barbecue sandwich= 0.90

0.75X1 + 0.45X2 + 0.90X3 ≤ 1500 (Cost)

Constraint 2: Space available = 3*4*16= 192 sq. feet

= 192*12*12= 27,648 sq. inches

Refilled oven before half time= 27,648*2= 55,296

Space required for pizza= 14*14= 196 sq. inches

Space required for pizza slice = 196/ 8 = 24.50 sq. inches

Space required for a hotdog=16

Space required for a barbecue sandwich = 25

24.50X1 + 16X2 + 25X3 ≤ 55,296 (Oven space)

Constraint 3: Selling at least as many slices of pizza X1, as hot dogs X2, and barbecue

sandwiches X3 combined

X1 ≥ X2 + X3 = X1 - X2 - X3 ≥ 0

X1 – X2 – X3 ≥ 0 (Selling X1, X2, X3)

Constraint 4: Selling at least twice as many hot dogs as barbecue sandwiches

X2/X3 ≥ 2 = X2 ≥2 X3 =X2 - 2 X3 ≥ 0

X2 - 2X3 ≥ 0 (Selling hot dogs & barbecue sandwich)

If the oven space required for a pizza slice was determined by dividing the total space required by a pizza, 14 x 14 = 196 in2 by 8, or approximately 24 in2 per slice. The total space available is the dimension of a shelf, 36 in. x 48 in. = 1,728 in2 multiplied by 16 shelves, 27,648 in2 which is multiplied by 2, the times before kickoff and halftime the oven will be filled = 55,296 in2.

Solution:

X1= 1,250 pizza slices

X2= 1,250 hot dogs

X3= 0 barbecue sandwiches

Z = $2,250

Julia should receive a profit of $2,250 for the first game....