R(p)=-1/4p^2+1500p

Question

suppose a company finds that the revenue, in dollars, from sales of a particular product is a function of the unit price p, in dollars, that it changes. the revenue R is given by:

 

R(p)=-1/4p^2+1500p

 

a. what unit price should be set to maximize revenue?

 

b. what is the maximum revenue?

 

Answer 1

Dear Sam,

 

First, think about what the function looks like. Does R(p) remind you of the form y = ax²+ bx + c? Good, so it's the quadratic equation for a parabola. Recall (from translations) that if the coefficient a is negative, then it's flipped over the x axis, i.e. resembles an upside down cup, sort of. 

 

Knowing that helps us to see that the maximum revenue (the y value) is at the vertex of the parabola, and the unit price here is p (the x value of the vertex). Think of it as the price per one unit of product, for example, a price for a football the company sells. In business, companies always look to maximize their profits.

 

So for part a, if the vertex is at (h, k) [but you might want to think of it as (p, R(p)) ], unit price = h = -b/2a. Plug in b=1500 and a=-1/4.

 

For part b, it's just your answer from part a plugged into the original function wherever you see a "p." 

 

Let me know if you get the answer. Good luck!

 

~Marina