Homework Help: math

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

a cable television firm presently serves 5000 households and charges $20 per month. A merketing survey indicates that each decrease of $1 in the monthly charge will result in 500 new costumers. Let R(x)denote the total monthly revenue when the monthly charge is x dollars.

Determine the revenue function R

math - Reiny, Monday, May 25, 2009 at 11:10pm
let 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 11:23pm
why 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 11:33pm
isn'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 11:12pm
The 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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