Expert: Paul Klarreich - 6/16/2006

Question
Hello Paul.  I'm having a very difficult time figuring out a certain calculus word problem involving revenue.  The subject is Calculus I.  I have already worked the problem at least 20 times by finding the demand function, and then plugging in the value of p(x).  Can you PLEASE help me out with this problem?  The problem is:

A manufacturer has been selling 1120 television sets a week at $380 each. A market survey indicates that for each $14 rebate offered to the buyer, the number of sets sold will increase by 196 per week. How large a rebate should the company offer the buyer in order to maximize its revenue?

I would greatly appreciate your help.  Thanks in advance!

Answer
Hi, Shawn,

You wrote:
Subject:  Calculus Revenue Problem
Question:  Hello Paul.  I'm having a very difficult time figuring out a certain calculus word problem

>> 'verbal' problem, please.  To my mind, a 'word' problem is something like:  John and Mary wrote essays.  John's essay had twice as many words as Mary's.......

involving revenue.  The subject is Calculus I.  I have already worked the problem at least 20 times by finding the demand function, and then plugging in the value of p(x).  Can you PLEASE help me out with this problem?  The problem is:

A manufacturer has been selling 1120 television sets a week at $380 each. A market survey indicates that for each $14 rebate offered to the buyer, the number of sets sold will increase by 196 per week. How large a rebate should the company offer the buyer in order to maximize its revenue?

I would greatly appreciate your help.  Thanks in advance!
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Start with a suitable variable or two.  Such as:

Let r = the amount of rebate offered.
   n = the number of sets sold.  (will be a function of r, according to your conditions.
   p = the price of the set. (Equal to 380 - r)

So it looks as if the increase in number of sets sold is  196/14, or 14 sets per dollar of rebate.  This might have been the confusing part.  When you are told '$r gives you so much increase in sales', get that as an increase per $1 of rebate.

Then the number sold, n = 1120 + 14r.
And the price per set is  p = 380 - r
So the revenue,  R = number sold * price per set.

R = (1120 + 14r)(380 - r)

That's your function to be maximized.  Multiply it out and it's a nice quadratic function with negative r^2 term, so it will have a nice maximum point.

I think you can take it from there.