# Thread: Maximizing Revenue!

1. ## 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 - . How many candy bars must be sold to maximize revenue?

I thought I could solve this problem by using implicit differentiation

2. Originally Posted by Matho
where p(x) = 39 - .
What is that supposed to mean? I dont get it, sorry

3. Originally Posted by Rapha
What is that supposed to mean? I dont get it, sorry
That is 39 - x/24

4. 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

Im still confused about

p(x) is the price charged for ONE candy bar, p(x) = 39 - x /24,

but x thousand candy bars will be sold

not sure about, maybe it is x*p(x)*1000, I dont know

5. Thank you so much! That really helps me to solve this problem