1. Mar 29, 2006

Aka

A large dealership had been selling new cars at $6000 over the factory price. Sales have been averaging 80 cars per mouth. Because of inflation, the$6000 markup is going to be increased. The marketing manager has determined that, for every $100 increase, there will be one less car sold each month. What should the new markup be in order to maximize the revenue? Let x rep the number of$100 increases
the equation for revenu is (#sold)(price)
so (80-x)(6000+100x)
then, when you expand it its -100x^2+2000x+480 000
next, solve using the quadratic formula
is everything right?

Thanks

2. Mar 29, 2006

arildno

Why would you find the values of x giving zero profit?

Find the MAXIMUM of the given profit function, not its roots.

3. Mar 29, 2006

Aka

^ok, thanks, so if my work abouve was correct, the answer should be a maximum of 490 000, when x=10

Last edited: Mar 29, 2006