# Maximizing Revenue!

• Nov 28th 2008, 02:28 PM
Matho
Maximizing Revenue!
If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in a certain city, where p(x) = 39 - http://cccs.blackboard.com/webct/Rel...les/f1q3g1.jpg. How many candy bars must be sold to maximize revenue?

I thought I could solve this problem by using implicit differentiation
• Nov 28th 2008, 06:48 PM
Rapha
Quote:

Originally Posted by Matho

What is that supposed to mean? I dont get it, sorry
• Nov 28th 2008, 06:51 PM
Matho
Quote:

Originally Posted by Rapha
What is that supposed to mean? I dont get it, sorry

That is 39 - x/24
• Nov 28th 2008, 07:02 PM
Rapha
Quote:

Originally Posted by Matho
That is 39 - x/24

Ok.

p(x) = 39 - x/24 is the price for a candy bar
you sell x candy bars, then your profit is

x*p(x) = 39x - x^2 / 24 =: f(x)

=> f'(x) = 39 - 2*x^1 / 24 = 39 - x/12

(f''(x) = - 1 /12 < 0 => Maximum )

f'(x) = 0

39 - x/12 = 0

// + x/12

39 = x/12

// * 12

x = 39 * 12 = 468

f''(x) < 0 => x = 468 Maximum

-----
x*p(x) = 39x - x^2 / 24 =: f(x)
------

f(468) = 468*(39 - 468 / 24) = 9126