# Mat540 Julia's Food Booth Case Problem

Topics: Optimization, Hot dog, Sausage Pages: 4 (934 words) Published: June 7, 2012
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

|Target Cell (Max) |  |  |  |  |  | |  | | | | | | |  |Cell |Name...