maximize quadratic profit fcn R(p) = -1/2p^2 + 1900p

[dbob85]

The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p it charges. If the revenue R is given by:

R(p) = -1/2p^2 + 1900p

(NOTE: "p^2" means "p squared".)

What unit price should be charged to maximize revenue? What is the maximum revenue?

How would I start this problem?

 

[stapel]

Since you mention that this is a quadratic function, I will guess that you actually mean the function to be:

. . . . .R(x) = -(1/2)p^2 + 1900p

Since you are asked for the maximum point of a negative quadratic, probably a good place to start would be finding the vertex.

Eliz.