# 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...

Please join StudyMode to read the full document