Math CentralQuandaries & Queries


Question from Chris, a student:

Lemon Motors have been selling an average of 60 new cars per month at $800 over the factory price. They are considering an increase in this markup. A marketing survey indicates that for every $20 increase, they will sell 1 less car per month. What should their new markup be in order to maximize income?

We have two responses for you

Hi Chris,

I think you want to maximize the income over the factory price. The income after paying the factory price to the manufacturer is the price charged over the factory price times the number sold. At the moment this is $8000 × 60. The intent is to increase the price some multiple of $20 per car. Let x be that multiple, and thus the new price will be $(800 + 20 x). The marketing survey says that for each x the number sold will decrease by one car and hence the number sold at this price will be (60 - x). Thus the income over the factory price will be

$(800 + 20 x) (60 - x)

Use the calculus you know to find the value of x that maximizes this expression.



Hi Chris.

A second way to solve the problem is to use your knowledge of parabolas rather than calculus.

y = (800 + 20 x) (60 - x)

is a quadratic equation. If you multiply this out and turn it into the y = a(x - p) + q "vertex graphing" form, then you can quickly determine the vertex of the equation. Since "a" is negative in your case, this is a parabola opening downwards ("concave down"). So the maximum y value occurs at the vertex.

Steve La Rocque.

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