Julia's Food Booth

Assignment #3: Julia's Food Booth

Hope Whaley

Professor: Dr. Daryl Brydie

Quantitative Methods - MAT 540

August 19, 2012

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

X1 – Pizza

X2 – Hot Dogs

X3 – Barbecue Sandwiches

Objective: Maximize profits = Profit at least $1,000.00 after each game.

Profit = Sell – Cost

Profit Function: Z = 0.75(X1) + 1.05(X2) + 1.35(X3)

Constraints and Cost:

Maximum fund available for the purchase = $1,500.00

Cost per pizza slice = $0.75 because she purchase each pizza for $6.00 and there are 8

slices per pizza.

Cost per hot dog = $0.45

Cost per sandwich = $0.90

Constraint = 0.75(X1) + 0.45(X2) + 0.90(X3) < 1500

Warming Oven:

Space available: 3 x 4 x 16 = 192 sq feet = 192 x 12 x 12 = 27,648 sq inches

Oven will be filled twice per game: 27,648 x 2 = 55,296

Space required for pizza: 14 x 14 = 196 sq inches

Space required for each pizza slice: 196/8 = 24.5 sq inches

Space required for each hot dog: 16 sq inches

Space required for each sandwich: 25 sq inches

Constraint: 24.5(X1) + 16(X2) + 25(X3) < 55,296

Julia can sell twice as many pizza slices(X1) as hot dogs(X2) and

sandwiches(X3), combined.

X1 > X2 + X3 = X2 – 2(X3) > 0

Julia can sell twice as many hot dogs(X2) as barbecue sandwiches(X3).

X2/X3 > 2 = X2 > 2(X3) = X2 – 2(X3) > 0

X1, X2, X3 >= 0 (Non-negativity constraint)

LPP Model:

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

Subject to 24.5(X1) + 16(X2) + 25(X3) < 55,296

0.75(X1) + 0.45(X2) + 0.90(X3) < 1,500

X1 – X2 – X3 > 0

X2 – 2(X3) > 0

X1 > 0, X2 > 0 and X3 > 0

Food Items: |Pizza |Hot Dogs |Barbecue | | | | | |Profit Per Item: |$0.75 |$1.05 |$1.35 | | | | | | | | | | | | | |...