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September 20, 2014

September 20, 2014

Posted by **Anonymous** on Monday, May 25, 2009 at 10:47pm.

Determine the revenue function R

- math -
**Reiny**, Monday, May 25, 2009 at 11:10pmlet the number of $1 decreases be n

then the monthly cost = 20-n

number of customers = 5000 + 500n

but you want 20-n = x

so n = 20-x

and 5000 + 500n

= 5000 + 500(20-x)

= 5000 + 10000 - 500x

= 15000-500x

so P(x) = x(15000 - 500x) or 15000 - 500x^2

(this is a very common question, but it is strangely worded.

Usually the question would be,

"What should be monthly charge for a maximum revenue ?" )

- math -
**Anonymous**, Monday, May 25, 2009 at 11:23pmwhy do u multiply everything by x at the end? and what is 15000-500x equal to? the number of costumers?

- math -
**Reiny**, Monday, May 25, 2009 at 11:33pmisn't the revenue = cost per month x number of customers?

didn't I define the "number of customers"

as 5000 + 500n

but x was 20-n which gave me n = 20-x

I then subbed that in 5000 + 500n

to get 15000 - 500x as I showed above step by step.

- math -
**PurduePete**, Friday, October 14, 2011 at 11:12pmThe above is all correct except for one mistake. x(15,000-500x) is not 15,000-500x^2 but rather 15,000x-500x^2. The x was not properly distributed in the original answer.

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