Question

The Camera Shop sells two popular models of digital SLR cameras (Camera A Price: 230, Camera B Price: 310). The sales of these products are not independent of each other, but rather if the price of one increase, the sales of the other will increase. In economics, these two camera models are called substitutable products. The store wishes to establish a pricing policy to maximize revenue from these products. A study of price and sales data shows the following relationships between the quantity sold (N) and prices (P) of each model: NA = 192 - 0.5PA + 0.25PB NB = 305 + 0.08PA - 0.6PB Construct a model for the total revenue and implement it on a spreadsheet. Develop a two-way data table to estimate the optimal prices for each product in order to maximize the total revenue. Vary each price from $250 to $500 in increments of $10.

Max profit occurs at Camera A price of $ _______

Max profit occurs at Camera B price of $ _______

Answer 1

Answer :-

Given data :-

Demand function for camera A and camera B are

NA = 192 - 0.5 PA + 0.25 PB

NB = 305 + 0.08 PA - 0.6 PB  

Here we need to find out the max profit occurs at Camera A price and max profit occurs at Camera B price.

Now, for revenue maximization problem, consider a Bertrand duopoly market.

Now consider total revenue,

from camera A :

TR = PA.NA

= PA( 192 - 0.5 PA + 0.25 PB )

TR = 192 PA - 0.5 PA^2 + 0.25 PA.PB -----------(A)

Now consider differentiation with respect to PA and equate it to zero., we get

--------(1)

from camera B :

TR = PB.NB

TR = PB ( 305 + 0.08 PA - 0.6 PB )

TR = 305 PB+ 0.08 PA. PB - 0.6 PB^2 -----------(B)

Now consider differentiation with respect to PB and equate it to zero., we get

----------(2)

Now multiply equation(a) with 0.08 , we get

--------(3)

Now by solving equations (3) and (2) we get

---------------------------------------------------------

  

Now from equation (a) we get,

  

  

  

  

Now from equation (A), By substituting PA and PB values, we get

TR = 192 (262.62) - 0.5 (262.62)^2 + 0.25(262.62)(271.69)

TR =50423.04 - 34484.63 + 17837.80

TR = $ 33776.214

max profit occurs at Camera A price of $ 33776.214

Now from equation (B), By substituting PA and PB values, we get

TR = 305 (271.69) + 0.08 (262.62).(271.69) - 0.6 (271.69)^2

TR =82865.45 + 5708.09 - 44289.27

TR = $ 44284.27

max profit occurs at Camera B price of $ 44284.27