The Camera Shop sells two popular models of digital cameras

The Camera Shop sells two popular models of digital cameras. 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:

   N_A = 190 - 0.7P_A + 0.35P_B

   N_B = 300 + 0.07P_A - 0.6P_B

Construct a model for the total revenue and implement it on a spreadsheet. Develop 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 $ .

Solution

N_A = 190 - 0.7P_A + 0.35P_B

   N_B = 300 + 0.07P_A - 0.6P_B

where N is the quantity sold and P the price

Revenue from car A = N_A*P_A = 190P_A-0.7P_A2+0.35P_AP_B

Revenue from car B = 300P_B + 0.07P_AP_B - 0.6P_B²

Total revenue R =  190P_A+300P_B-0.7P_A2-0.6P_B²+0.42P_AP_B

Let P_A,P_B be denoted by x and y

R(x,y) = 190x+300y-0.7x²-0.6y²+0.42xy

x varies from 250 to 500 and y 250 to 500

Maximum when both are equal to 500

Using partial derivatives

R(x,y) = 190x+300y-0.7x²-0.6y²+0.42xy

R_x = 190-0.14x+0.42y

R_y = 300-0.12y+0.42x

Set both to 0 and solve for x and y

Multiply I equation by 3

570-0.42x+1.26y =0 and

300+0.42x-0.12y=0

Adding 870 = 1.14y

y = 763.14

300-0.12(763.14)+0.42x=0

x = 496.24

i.e. when x = 496 and y =763 revenue is maximum.

Max profit occurs at Camera A price of $ 496

Max profit occurs at Camera B price of $ 763