X2 = Hot Dogs X3 = BBQ Sandwiches The objective is to maximize profit. maximize Z= 0 .75X1+1.05X2+1.35X3 Subject to: 0.75X1+1.05X2+1.35X3≤1‚500(Budget) 24X1+16X2+25X3≤55‚296in2 (Oven space) X1≥X2+X3 X2X3≥2.0 X1‚ X2‚ X3≥0 Julia’s Food Booth Food items: Pizza Hot Dogs Barbecue Profit per item: 0.75 1.05 1.35 Constraints: Available Usage Left over Budget ($) 0.75 0.45 0.90 1‚500 1‚500.00 0 Oven space (sq. in.) 24 16 25 55‚296 50‚000.00 5296 Demand
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Formulation of the LPP Suppose Julia stock X1 numbers of pizza slices‚ X2 numbers of hot dogs and X2 numbers of sandwiches. Constraints: 1. On the 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 during half time. Thus total space available = 27648 x 2 = 55296 Space required for pizza = 14 x 14 = 196 sq. inches Space required for pizza slice = 196/ 8 = 24.5 sq. inches Total Space required: 24.5 X1 + 16 X2
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Julia Food Booth Introduction Julia is planning to lease a food booth outside the Tech Stadium at Home Football games to finance her last year education with all the games go sold out. The rent for the booth per game is $ 1000. Julia will sell slices of Cheese Pizza‚ Hot Dogs and Barbecue Sandwiches which are acclaimed to be the most popular so these are the three products she has chosen to sell at the home games football stadium. The rent for oven is $ 600 for six home games‚ which
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Julia’s Food Booth (A) Formulate and solve an LP model for this case The objective here is to maximize the profit. Profit is calculated for each variable by subtracting cost from the selling price. The decision variables used are X1 for pizza slices‚ X2 for hotdog‚ and X3 for BBQ sandwich. X1 (pizza) X2 (hotdog) X3 (sandwich) Sales Price 1.50 1.50 2.25 Cost 0.75 0.45 0.90 Profit 0.75 1.05 1.35 *For Pizza Slice: Cost/Slice = $6/8 = $0.75 cost per slice Maximize Z =
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a) 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
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Running Head: JULIA’S FOOD BOOTH Assignment #3: Case Problem "Julia’s Food Booth" Mat540 Quantitative Methods August 22‚ 2012 Julia’s Food Booth (A) Formulate and solve a L.P. model for this case. Variables: Pizza - X1 $1.33 $1.50 14 inches Hot Dogs - X2 $0.45 $1.50 16 square inches Barbecue - X3 $0.90 $2.25 25 square inches Maximize Z= $0.75x1‚ 1.05x2‚ 1.35x3 Subject to: $0.75x1 + $0.45x2 + $0.90x3 ≤ $1‚500 24x1 + 16x2 + 25x3 ≤
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Julia Booth Case Study Angela Walker June‚ 06‚ 2013 Dr. Carl Tucker field In this case study‚ I must create a linear programming model for Julia that will help her to decide whether or not she need to lease a booth. The three products or variable we must consider for this booth are pizza‚ hotdogs‚ and barbecue sandwiches. In this model‚ x 1 equal the number of pizza slices Julia should purchase‚ x2 equals the number of hotdogs Julia should purchase‚ and x3 equal the number of barbecue sandwiches
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Julia’s Food Booth Case Problem MAT 540- Quantitative Methods February 23‚ 2013 (A) Formulate and solve an L.P. model for this case. The following variables were be used: X1 = Slices of Pizza X2 = Hot Dogs X3 = BBQ Sandwiches The objective is to maximize profit. maximize Z= 0 .75X1+1.05X2+1.35X3 Subject to: 0.75X1+1.05X2+1.35X3≤1‚500 (Budget) 24X1+16X2+25X3≤55‚296in2 (Oven Space) X1≥X2+X3 X2X3≥2.0 X1‚ X2‚ X3≥0 (B) Evaluate the prospect of borrowing money before the first
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Julia’s Food Booth Case Problem Assignment 3 Max Z =Profit1x1+ Profit2x2+ Profit3x3 A - Formulation of the LP model x1 - number of pizza slice x2 - number of hot dogs x3 - number of barbecue sandwiches Constraints Cost Maximum fund available for food = $1500 Cost per pizza $6 ÷08 (slices) = $0.75 Cost for a hot dog = $0.45 Cost for a barbecue sandwich = $0.90 Constraint: 0.75x1+0.45x2+0.90x3 ≤1500 Oven space Space available 16.3.4.2 = 384ft^2 384
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Assignment #3: JULIA`S FOOD BOOTH A. Julia would make $1150 profit after paying all expenses after the first game. And then she would make $1721.22 for the rest of the games. Since she will be clearing her number of $1000 profit per game‚ she should lease the booth. B. If she borrows money from a friend she would increase her profit. She would borrow $380.82 from a friend and she will make $571.22 more profit. The factor that constraints her from borrowing even more money is the total
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