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?

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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
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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revenue= the price* quantity, let's g(p) be a new function, represent revenue \[g(p)=pf(p)=180p-0.3p^3\]
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:
\[g'(p)=180-0.9p^2=0 \implies p^2=200 \implies p=\pm10\sqrt2 \] 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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I hope that makes sense to you
at p = 0
^^ 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 :)
thanks
you're welcome
sorry the last step it's \[p=5\sqrt2\] sorry for the typo mistake
Try www.aceyourcollegeclasses.com
it is P^2=20 ---> p=5(sqrt2)
what's wrong with me?? :@ the first answer is right.. lol
the initial problem is f(p)=180-0.3p^2

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