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Julia's Food Booth Case

Julia's Food Booth Case - Space is the factor which...

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Julia’s Food Booth 1 Assignment #3- Julia’s Food Booth MAT 540 Quantitative Methods Dr. Vargha Azad June 5, 2011 Julia’s Food Booth a)
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Julia’s Food Booth 2 P=PIZZA H=HOTDOGS S=BARBEQUE SANDWHICHES The L.P. for this case is, Maximize: P= 0.75P +1.05H +1.35S Subject to: 24.5p +16h +25s ≤ 55296 0.75p + 0.45h+0.90s ≤ 1500 P – h – s ≥ 0 H - 2s ≥ 0 P , h , s ≥ 0 The optimal solution occurs when P=1250 h-1250 s=0 Profit =$2250 Since the booth costs $1000 to lease per game, and the oven is $100 per game, then Julia's overall profit is, P = 2250 - 1100 = $1150 Hence it is worth leasing the booth. b) The shadow price for the budget is 1.50 and allowable increase is 138.4 Therefore, each dollar added to the budget will increase profit by $1.50 with a maximum increase of $138.40. Therefore the maximum amount Julia can borrow is $138.40 which will produce an additional profit of 138.4 x 1.5 = $207.60.
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Unformatted text preview: Space is the factor which constrains Julia from borrowing more money. c) If Julia feels she needs help than $100 would still keep her above the $1000 dollar profit margin. Therefore, if Julia cannot handle the work load alone it would be advisable for her to hire some help. If she is able to handle the work load herself, then she could keep that $100 for herself. d) The biggest uncertainty in this model is demand. Although Julia may have a good idea of what people will buy and not buy during the game, the demand can shift from game to game and is not always constant. Hence, if the demand changes then the solution to the linear programming will change, and may affect her ability to make a profit greater than $1000....
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