## anonymous 5 years ago 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?

1. anonymous

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

2. anonymous

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:

3. anonymous

$g'(p)=180-0.9p^2=0 \implies p^2=200 \implies p=\pm10\sqrt2$ 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.

4. anonymous

I hope that makes sense to you

5. anonymous

at p = 0

6. anonymous

^^ 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 :)

7. anonymous

thanks

8. anonymous

you're welcome

9. anonymous

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

10. anonymous

Try www.aceyourcollegeclasses.com

11. anonymous

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

12. anonymous

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

13. anonymous

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