Chapter 2.1, Problem 39E

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Chapter
Section

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

# Website Profit The latest demand equation for your gaming website, www.mudbeast.net, is given by q = − 400 x + 1 , 200 where q is the number of users who log on per month and x is the log-on fee you charge. Your Internet provider bills you as follows:Site maintenance fee: $20 per monthHigh-volume access fee: 50 ¢ perlog-onFind the monthly cost as a function of the log-on fee x. Hence, find the monthly profit as a function of x, and determine the log-on fee you should charge to obtain the largest possible monthly profit. What is the largest possible monthly profit? To determine To calculate: The monthly cost as a function of the log-on fee x for the given demand equation, q=400x+1200 Where q is the number of users who log-on per month and x is the charged log-on fee. Also find the monthly profit as a function of x and determine the log-on fee that would lead to largest monthly profit. Explanation Given Information: The provided linear demand equation, q=400x+1200 Where q is the number of users who log-on per month and x is the charged log-on fee. The internet provider bills as follows: Site maintenance fee:$20

per month

High Volume access fee: 50

cents

Formula used:

The relationship between Revenue and price is,

Revenue=(Price)(Demand)R(x)=(x)(q)

The relationship between Revenue and profit is,

Profit=RevenueTotalCostP(x)=R(x)C

Calculation:

Consider the given data,

The provided linear demand equation,

q=400x+1200

Where q

is the number of users who log-on per month and x is the charged log-on fee. The internet provider bills as follows:

Site maintenance fee: $20 per month High Volume access fee: 50 cents Convert the given high volume access into dollars, 1cent=$1100 50cents=$50100 =$0.5

Total cost is given by,

Total cost=Site maintenance fee +(High volume acces fee)(Number of users)

Substitute 400x+1200 for q

in equation C=20+(0.5)(q):

C=20+(0.5)(400x+1200)=20200x+600=200x+620

Therefore, total monthly Cost as a function of log-on fee is, C=200x+620

Consider the given demand equation,

q=400x+1200

The relationship between Revenue and Price is,

Revenue=(Price)(Demand)R(x)=(x)(q)

Substitute 400x+1200 for q

in Revenue equation:

R(x)=(x)(q)=x(400x+1200)=400x2+1200x

Therefore, the total Revenue R

as a function of the log-on fee x for the given demand equation is R(x)=400x2+1200x.

Consider the given Revenue equation,

R(x)=400x2+1200x

The relationship between Revenue and profit is,

Profit=RevenueTotalCostP(x)=R(x)C

Substitute 400x2+1200x for R(x)

and 200x+620 for C equation P(x)=R(x)C

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