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 + 25 X3
Constraint: 24.5 X1 + 16 X2 + 25 X3 ≤ 55296

2. On the cost: Maximum fund available for the purchase = $1500 Cost per pizza slice = 6/8 = $0.75
Funds required: 0.75 X1 + 0.45 X2 + 0.90 X3
Constraint: 0.75 X1 + 0.45 X2 + 0.90 X3 ≤ 1500

3. On demand:
She can sell at least as many slices of pizza as hot dogs and barbecue sandwiches combined Constraint: X1 ≥ X2 + X3 [pic]X1 - X2 - X3 ≥ 0

She can sell at least twice as many hot dogs as barbecue sandwiches X2/X3 ≥ 2 (at least twice as many hot dogs as barbeque sandwiches) X2 ≥2 X3 [pic] X2 - 2 X3 ≥ 0

Objective Function (Profit):
Profit on pizza slice = $1.50 - $0.75 = $ 0.75
Profit on hot dog = $1.50 – 0.45 = $1.05
Profit on sandwich = $2.25 - $0.90 = $1.35
Profit function: Z = 0.75 X1 + 1.05 X2 + 1.35 X3
A) 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
Solution
Solution of the LPP by using Excel Solver gives the following Report

...Julia’sFoodBoothCaseProblem
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...

...Julia’sFoodBoothCaseProblem
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:...

...(A) Formulate and solve an L.P. model for this case
Variable Food Cooking Area
x1 Pizza Slice 24in sq
x2 Hot Dogs 16in sq
x3 BBQ Sandwiches 25in sq
*The oven space required for a pizza slice is calculated by dividing the total area arequired for a whole pizza by the number of slices in a pizza 14 x 14 = 196 in2, by 8, or approximately 24 in2 per slice. The total space available is the dimension of a shelf, 36 in. x 48 in. = 1,728 in2, multiplied by 16...

...Julia’sFoodBooth
Julia Robertson is a senior at Tech, and she's investigating different ways to finance her final year at school. She is considering leasing a foodbooth outside the Tech stadium at home football games. Tech sells out every home game, and Julia knows, from attending the games herself, that everyone eats a lot of food. She has to pay $1,000 per game for a booth, and the...

...A. Formulate a linear programming model for Julia that will help you to advise her if she should lease the booth.
Let, X1 =No. of pizza slices,
X2 =No. of hot dogs,
X3 = No. of barbeque sandwiches
* Objective function co-efficient:
The objective is to maximize total profit. Profit is calculated for each variable by subtracting cost from the selling price.
For Pizza slice, Cost/slice=$4.5/6=$0.75
| X1 | X2 | X3 |
SP | $1.50 | $1.60 | $2.25 |
-Cost | 0.75 |...

...Assignment #3: Julia’sFoodBooth
Quantitative Methods 540
Buddy L. Bruner, Ph.D.
Shirley Foster
11/25/2012
Assignment 3: Caseproblem “Julia’sFoodBooth” Page 1
A. Julia Robertson is making an allowance for renting a foodbooth at her school. She is seeking ways to finance her last year and believed that a...