# Julia's Food Booth Case Problem

Topics: Optimization, Mustard, Hamburger Pages: 4 (893 words) Published: March 11, 2013
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
384.144=55296 in ^2
Space required for pizza: 14.14 = 196 ^2 inches
Space for slice of pizza; 196 ÷8 = 24.50 in ^2
Space for hot dog: 16 in ^2
Space for barbecue = 25 in ^2
Constraint 24.50x1+16x2+25x3 ≤55296
Julia can sell at least as many slice of pizza (x1) as hot dogs (x2) and Barbecue sandwiches (x3) combined. x1-x2-x3≥0
Julia can sell at least twice as many hot dogs as Barbecue sandwiches +x2-2x3≥0
Non negative constraint
x1,x2,x3≥0
Objective Function
| SELL| COST| PROFIT|
Pizza slice (x1) | \$1.50 | \$0.75 | \$0.75| Hot dog (x2) | \$1.50 | \$0.45 | \$1.05| Barbecue Sandwich (x3) | \$2.25 | \$0.90 | \$1.35| Profit = Sell - Cost

Max Z=0.75x1+1.05x2+1.35x3
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 -answer in QM for Windows solution
Based on the QM for Windows solution the optimum solution:
Pizza (X1) = 1250; Hotdog s(X2) = 1250 and Barbecue sandwiches (X3) = 0 Optimal solution value 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| \$ (1,000.00)|
Warming Oven 600 for total of 6 home games 600/6 =100| \$ (100.00)| Profit for the 1st Game| \$ 1,150.00...