Another Quadratic Equation Question

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shanieO

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Mar 9, 2006
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Need some help with this question please.

A small corner store owner decides to advertise on local radio in order to help stimulate his sales. Radio station CFAL charges him a flate rate of $99, plus $20 per day for advertising. From the onlsaught of the advertising the revenue increase can be described as (120-x) dollars per day where x is the number of days the ad is run. He plans to run the ad for 60 consecutive days.
a) How many days must the ad run until the increased revenue just covers the cost of the advertising?
b) How many days should he let the ad run in order to maximize his profit from the ad campaign?

I have R(x) = (120x-x^2) P(x) = R(x) - C(x)
= (120x-x^2) - (20x + 99)
= -x^2 + 100x -99

Now I am lost as to what to do next.

Thanks for any help I can get.
 
For part a, you are close, but you are ignoring his decision to run it for 60 days. That makes the cost 99+20*60. That is what he must recoup.
x(120-x)=1299

Your equation applies to part b and is really asking how many days SHOULD he have been running the ad to maximize profit. You want to find the vertex at x=-b/2a.
 
My take on the problem is that he PLANS to run the ad 60 days, but
that he may at anytime cancel. In that case the quadratic you found
is the correct one.
a) Solve for R(x)=0; two answers , one way past the 60day deadline.

b) Find the vertex (h,k) to R(x) to determine how many days to run the ad
and how much revenue will be generated.

Good Luck!
 
So its a matter of opinion, but since the 60 days is part of the question I assume it is part of the answer and cannot be ignored. I'm sticking to my interpretation.
----------------
Gene
 
If he runs the ad for 60 days, as planned, it will cost
99+20*60=1299.
The revenue is, as you said, x(120-x). When they are equal he breaks even hence
120x-x²=1299
Where are you lost?
---------------
Gene
 
Gene:
Sorry, I should have paid more attention. I did get the first part I was meaning the second part.

x=-b/2a x=120/2(?) What would "a" be?

Thanks

[/i]
 
In your other post robbwrr defined the general quadratic equation as ax²+bx+c=0
a and b refer to that. Look over what he said.
You are solving
-1x^2 + 100x -99 = 0 so
a=-1, b= 100 and the vertex is at
x = -b/2a, a shortcut well worth memorizing.
 
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