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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?

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  1. anonymous
    • 5 years ago
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    revenue= the price* quantity, let's g(p) be a new function, represent revenue \[g(p)=pf(p)=180p-0.3p^3\]

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    \[g'(p)=180-0.9p^2=0 \implies p^2=200 \implies p=\pm10\sqrt2 \] 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
    • 5 years ago
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    I hope that makes sense to you

  5. anonymous
    • 5 years ago
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    at p = 0

  6. anonymous
    • 5 years ago
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    ^^ 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
    • 5 years ago
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    thanks

  8. anonymous
    • 5 years ago
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    you're welcome

  9. anonymous
    • 5 years ago
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    sorry the last step it's \[p=5\sqrt2\] sorry for the typo mistake

  10. anonymous
    • 5 years ago
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    Try www.aceyourcollegeclasses.com

  11. anonymous
    • 5 years ago
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    it is P^2=20 ---> p=5(sqrt2)

  12. anonymous
    • 5 years ago
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    what's wrong with me?? :@ the first answer is right.. lol

  13. anonymous
    • 5 years ago
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    the initial problem is f(p)=180-0.3p^2

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