revenue word problem

[arman0077]

hi
I solved this question
A cable television firm presently serves 8000 households and charges $50 per month. A marketing
survey indicates that each decrease of $5 in the monthly charge will result in 1000 new customers. Let R(x) denote the
total monthly revenue when the monthly charge is x dollars.
(a) Determine the revenue function R.
(b)find the value of x that results in maximum monthly revenue.
in this way
[MATH] R(x) = (8000+1000x)(50-5x) [/MATH]and for b
[MATH]8000+1000x=0 -> x = -8[/MATH][MATH]50 - 5x = 0 -> x = 10[/MATH][MATH] \frac{10+-8}{2}=1[/MATH]
so it was my answer but the answer to the question was
a)
[MATH]R(x) = 200x(90 - x)[/MATH]and b)
45 $
I check my answer multiple times but couldn't figure out where am I making the mistake the only thing that I came up with was that the root of my variable x is very illogical with its definition that is the monthly charge but couldn't go any further

 

[wuti]

hi, i think the answer is correct. so after adjustment price is x, which is decreased from original price 50, therefore, you must lower your price by (50-x), which will give you (50-x)/5*1000 more customers. so total customers will be 8000+(50-x)/5*1000, times price x will give you revenue. so R=[8000+(50-x)/5*1000]*x=(90-x)*x*200. for (b) just use formula.

 

[lev888]

Your x is not the monthly charge as stated in the problem. You need to correct your R(x).