MAT assignment #3

MAT assignment #3 - Running head: JULIA'S FOOD BOOTH...

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Running head: JULIA’S FOOD BOOTH Julia’s Food Booth Strayer University Quantitative Methods – MAT 540 Dr Alexander Thomas November 19, 2011
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A. Formulate a linear programming model for Julia that will help you to advise her if she should lease the booth. Formulate the model for the first home game. Explain how you derived the profit function and constraints and show any calculations that allow you to arrive at those equations. X1 – No. of pizza slices X2 – No. of hotdogs X3 – No. of barbeque sandwiches. Constraints: Space available = 3 *4 * 16 = 192 sq. feet = 192 * 12 * 12 =27648 sq. inches The oven will be refilled during half time. Total space available = 27648 * 2 = 55296 in square Space required for pizza = 14 * 14 = 196 sq. inches Space required for pizza slice = 196/ 8 = 24.5 sq. inches Total Space required: 24.5 X1 + 16 X2 + 25 X3 Constraint: 24.5 X1 + 16 X2 + 25 X3 ≤ 55296 (oven space) Julia’s Budget = $1500 Cost per pizza slice = 6/8 = $0.75 Funds required: 0.75 X1 + 0.45 X2 + 0.90 X3 Constraint: 0.75 X1 + 0.45 X2 + 0.90 X3 ≤ 1500 Sell as many slices of pizza as hot dogs and barbecue sandwiches combined Constraint: X1 ≥ X2 + X3 X1 - X2 - X3 ≥ 0 Sell as twice as many hot dogs as barbecue sandwiches X2/X3 ≥ 2.0 Constraint: X2 ≥2 X3 X2 - 2 X3 ≥ 0 Profit: Profit on pizza slice = $1.50 - $0.75 = $ 0.75 Profit on hot dog = $1.50 – 0.45 = $1.05 Profit on sandwich = $2.25 - $.90 = $1.35 Profit function: Z = 0.75 X1 + 1.05 X2 + 1.35 X3 Model: Subject to: Maximize the profit Z = 0.75 X1 + 1.05 X2 + 1.35 X3
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Total Space Required: 24.5 X1 + 16 X2 + 25 X3 ≤ 55296 Budget:
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This note was uploaded on 02/10/2012 for the course MAT 540 taught by Professor Dralexander during the Spring '11 term at Strayer.

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MAT assignment #3 - Running head: JULIA'S FOOD BOOTH...

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