Given f(p)=180-0.3p^2, where p is the price in dollars and f(p) is the number of items sold. At what price will maximum revenue be generated?

Question

anonymous · Answer

revenue= the price* quantity,
let's g(p) be a new function, represent revenue
$$g(p)=pf(p)=180p-0.3p^3$$

anonymous · Answer

now just take the derivative of g(p), set it equal to zero, and find the value of p (the price) for maximum revenue as follow:

anonymous · Answer

$$g'(p)=180-0.9p^2=0 \implies p^2=200 \implies p=\pm10\sqrt2 $$
price can never be a negative value, therefore we will take only the positive value, which is $$p=10\sqrt2$$
that's the price at which maximum revenue will be generated.

anonymous · Answer

I hope that makes sense to you

anonymous · Answer

at p = 0

anonymous · Answer

^^
p=0 is the price to maximize f(p) clearly, which is the number of sold items. you will never get maximum revenue selling things for free :)

anonymous · Answer

thanks

anonymous · Answer

you're welcome

anonymous · Answer

sorry the last step it's
$$p=5\sqrt2$$
sorry for the typo mistake

anonymous · Answer

Try www.aceyourcollegeclasses.com

anonymous · Answer

it is P^2=20 ---> p=5(sqrt2)

anonymous · Answer

what's wrong with me?? :@ the first answer is right.. lol

anonymous · Answer

the initial problem is f(p)=180-0.3p^2