The monthly revenue R achieved by selling x wristwatches is

The monthly revenue R achieved by selling x wristwatches is figured to be R(x) = 15x-0.2x^2. The monthly cost C of selling x wristwatches is C(x) =32x+1750. How many wristwatches must the firm sell to maximize revenue? What is the maximum revenue? Profit is given as P(x) = R(x)-C(x). Why does this definition make sense? Find the profit function. How many wristwatches must the firm sell to maximize profit? What is the maximum profit? Provide a reasonable explanation as to why the answers from parts a and c differ. Explain why a quadratic function is a reasonable model for revenue.

Solution

a) R(x) = 75x -0.2x^2

maximum revenue is found at R\'(x) =0 ; 75 -0.4x=0

x = 187.5 = 187 watches

Maximum reveue at x= 187 : R(187) = $7031.2

b) Profit = R(x) - C(x) = 75x - 0.2x^2 - (32x +1750) =-0.2x^2 + 43x -1750

c) Maximium profit is found at P(x)\'=0

-0.4x +43 =0 ----> x = 107.5 = 107 watches

P(107) = -0.2*107^2 +43*107 -1750 = $ 561.2

d) In part a ) we found maximum revenue i.e. sales of watches, however we are looking at achieving the maximum profit.Both are diffrent things

Revenue is a quadratice function because it is a product of demand price ( linear function of x) and number of items sold ( x).So, it would be a quadraric function