Bearcat Fans Clothing Store is planning for the upcoming basketball season. They
ID: 2930390 • Letter: B
Question
Bearcat Fans Clothing Store is planning for the upcoming basketball season. They have 2 shirts that they will be selling that look like the warm-up shirts that the Men's Basketball Team will be wearing. They have an screen print shirt (S) and an embroidered shirt (E) that they will be selling. Demand for the two shirts depends on selling price of both shirts. The demand for these two shirts are are follows: (DS is demand for the screen print shirt, PS is the selling price for the screen print shirt, DE is demand for the embroidered shirt, PE is the selling price for the embroidered shirt):
DS = 150 - 10PS + 1.8PE
DE = 525 +3.4PS - 19PE
The store wishes to determine the selling price that maximizes revenue for these two shirts. Develop a nonlinear optimization model to determine the optimal selling prices.
The Mathematical Model for the problem is
Maximize PS*DS + PE*DE - Which could be written PS*(150 - 10PS + 1.8PE )+ PE*(525 +3.4PS - 19PE)
Subject to: PS, PE, DS, DE >= 0
A. What is the optimal selling price for the screen print shirt? (round your answer to 2 decimal places)
B. What is the optimal selling prices for the embroidered shirt? (round your answer to 2 decimal places)
C. What is the maximum revenue? (round your answer to 2 decimal places)
Explanation / Answer
DS = Demand for the screen print shirt
PS = Selling price for the screen print shirt
DE = Demand for the embroidered shirt
PE = Selling price for the embroidered shirt
Here as per question
DS = 150 - 10PS + 1.8PE
DE = 525 +3.4PS - 19PE
We have to maximize
Total Revenue TR = PS * DS + PE * DE
TR = PS * (150 - 10PS + 1.8PE ) + PE*(525 +3.4PS - 19PE)
TR = 150 PS - 10PS2 + 1.8 PE .PS + 525 PE + 3.4 PE.PS - 19 PE2
TR = - 10PS2 - 19 PE2 + 5.2 PE.PS + 150 PS + 525 PE
Subject to: PS, PE, DS, DE >= 0
TO find optimum price of screen print shirt and embroidered shirt.
so first we would differentiate it with PS to find optimum price of print shirt and differentiate it with PE to find optimum price of embroidered shirt
d(TR)/ d(PS) = -20 PS + 5.2 PE + 150 ...(i)
d(TR)/ d(PE) = -38 PE + 5.2 PS + 525 ..(ii)
so both values would be zero for optimum price of print shirt and embroidered shirt.
-20PS + 5.2 PE + 150 = 0
-38 PE + 5.2 PS + 525 = 0
solving both equations simultaneously
PS = $11.50 , PE = $ 15.39
(a) SO Optimum price of Print shirt = $ 11.50
(b) Optimum Price of Emproided shirt = $ 15.39
(c) Tota revenue = - 10PS2 - 19 PE2 + 5.2 PE.PS + 150 PS + 525 PE
= - 10 * 11.502 - 19 * 15.392 + 5.2 * 11.50 * 15.39 + 150 * 11.50 + 525 * 15.39
= $ 4902.38
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