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Julia's Food Booth

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Case Problem Julia's Food Booth
Assignment #3 Case Problem Julia's Food Booth

A: Formulation of the LP Model

X1(Pizza), X2(hotdogs), X3(barbecue sandwiches)

Constraints:

Cost:

Maximum fund available for the purchase = $1500

Cost per pizza slice = $6 (get 8 slices) =6/8 = $0.75

Cost for a hotdog = $.45

Cost for a barbecue sandwich = $.90

Constraint: 0.75X1 + 0.45X2+ 0.90(X3) ≤ 1500

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 before half time- 27648 x 2 = 55296

Space required for pizza = 14 x 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

Constraint: 24.50 (X1) + 16 (X2) + 25 (X3) ≤ 55296

Constraint:

Julia can sell at least as many slices of pizza(X1) as hot dogs(x2) and barbecue sandwiches (X3) combined

Constraint: X1 ≥ X2 + X3 = X1 - X2 - X3 ≥ 0

Julia can sell at least twice as many hot dogs as barbecue sandwiches

X2/X3 ≥ 2 = X2 ≥2 X3 =X2 - 2 X3 ≥ 0

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

Objective Function (Maximize Profit):

Profit =Sell- Cost

Profit function: Z = 0.75 X1 + 1.05 X2 + 1.35 X3

LPP Model:

Maximize Z = 0.75 X1 + 1.05 X2 + 1.35 X3

Subject to 24.5 X1 + 16 X2 + 25 X3 ≤ 55296

0.75 X1 + 0.45 X2 + 0.90 X3 ≤ 1500

X1 - X2 - X3 ≥ 0

X2 - 2 X3 ≥ 0

X1≥ 0, X2≥ 0 and X3 ≥0

Solve the LPM

Based on the excel solution the optimum solution:

Pizza (X1) = 1250; Hotdogs(X2) = 1250 and Barbecue sandwiches (X3) = 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 |$ 2,250.00 |
|Booth Rent per game |$

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