MAT500 Assignment 3 - Running head: CASE PROBLEM JULIAS...

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Running head: CASE PROBLEM “JULIA’S FOOD BOOTH” 1 Case Problem “Julia’s Food Booth” Strayer University MAT540 – Quantitative Methods August11, 2011
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CASE PROBLEM “JULIA’S FOOD BOOTH” 2 Case Problem “Julia’s Food Booth” Formulate and Solve an L.P. Model for this Case Following is the model formation and solution steps for this case. Decision Variables This problem includes three decision variables, representing the number of each food item to purchase: x 1 = number of pizza slices to purchase x 2 = number of hot dogs to purchase x 3 = number of barbeque sandwiches to purchase Objective Function The objective of Julia’s food booth is to maximize the total profit. The total profit is the sum of the individual profits gained from pizza slices, hot dogs and barbeque sandwiches. Profit derived from a pizza slice = $1.50 – ($6/8 slices) = $1.50 - $0.75 = $0.75 Profit derived from a hot dog = $1.50 - $0.45 = $1.05 Profit derived from a barbeque sandwich = $$2.25 - $0.90 = $1.35 Thus, total profit, which we will define symbolically as Z, can be expressed mathematically as $0.75x 1 + $1.05x 2 + $1.35x 3 . The objective function can be written as: maximize Z = $0.75x 1 + $1.05x 2 + $1.35x 3 . where Z = total profit for the first home game $0.75 x 1 = profit from pizza slices $1.05x 2 = profit from hot dogs $1.35x 3 = profit from barbeque sandwiches
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CASE PROBLEM “JULIA’S FOOD BOOTH” 3 Model Constraints The first constraint is budget constraint. Money available for purchase is $1,500. Cost for the purchase of per pizza slice is $0.75 (=$6/8 slices). Cost for the purchase of per hot dog is $0.45. Cost for the purchase of per barbeque sandwich is $0.90. Total cost for the purchase pizzas, hot dogs, and barbeque sandwiches is $0.75x 1 + $0.45x 2 + $0.90x 3. Therefore, the budget constraint is expressed as: $0.75x 1 + $0.45x 2 + $0.90x 3 <= $1,500 The second constraint is space constraint. The oven has 16 shelves, and each shelf is 3 feet by 4 feet. One foot equals to 12 inches. The total space (of 16 shelves) available =3*4*16=192 sq. feet =192*12*12=27,648 sq. inches. Since the oven can be refill two times during a game, the total space available is double. Thus, the total space available = 27,648 * 2= 55,296 sq. inches. On the other hand, space required for a pizza=14*14=196 sq. inches. Since each pizza contains 8 slices, space required for a pizza slice 196/8=24.50 sq. inches. Total space taken by pizzas, hot dogs, and barbeque sandwiches is 24.50x 1 + 16x 2 + 25x 3. Thus, the oven
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This note was uploaded on 11/15/2011 for the course ACCOUNTING MAT540 taught by Professor Georgemaruschock during the Summer '11 term at Strayer.

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MAT500 Assignment 3 - Running head: CASE PROBLEM JULIAS...

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