# Mat540 Julia's Food Booth Case Problem

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...

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