anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
revenue= the price* quantity, let's g(p) be a new function, represent revenue \[g(p)=pf(p)=180p-0.3p^3\]
anonymous
  • 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:
anonymous
  • 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.

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anonymous
  • anonymous
I hope that makes sense to you
anonymous
  • anonymous
at p = 0
anonymous
  • 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 :)
anonymous
  • anonymous
thanks
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
sorry the last step it's \[p=5\sqrt2\] sorry for the typo mistake
anonymous
  • anonymous
Try www.aceyourcollegeclasses.com
anonymous
  • anonymous
it is P^2=20 ---> p=5(sqrt2)
anonymous
  • anonymous
what's wrong with me?? :@ the first answer is right.. lol
anonymous
  • anonymous
the initial problem is f(p)=180-0.3p^2

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