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Math Help - Max Value of: Ln(x)/(x^2)

  1. #1
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    Max Value of: Ln(x)/(x^2)

    How do you find the maximum value of (ln(x))/x^2

    First i derived: That gave me: (1-2ln(x))/x^3
    Now i figure x' is zero at: x = 0
    or ln(x) = 0,5 = e^0,5

    But the correct answer is suppose to be 1/(2e)

    What am i doing wrong?
    With Kind regards Henry from Sweden
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  2. #2
    Moo
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    Hello,

    the derivative at x=0 isn't 0... it's undefined (because there's x in the denominator)

    but your answer of x_{\max}=e^{1/2} is correct.

    Now what is f(x_{\max}), where f(x)=\frac{\ln(x)}{x^2} ? This is the maximum value of the function. What you found is just the point where the max is attained !

    Oh, you also have to check that it's a max, that is to say that f''(x_{\max})<0
    Last edited by Moo; December 29th 2009 at 07:23 AM. Reason: you're -> your
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    Quote Originally Posted by Henryt999 View Post
    How do you find the maximum value of (ln(x))/x^2

    First i derived: That gave me: (1-2ln(x))/x^3
    Now i figure x' is zero at: x = 0
    or ln(x) = 0,5 = e^0,5

    But the correct answer is suppose to be 1/(2e)

    What am i doing wrong?
    With Kind regards Henry from Sweden
    there is a global maximum at \sqrt{e}.
    \sqrt{e} is the root of your first derivative.
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  4. #4
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    Quote Originally Posted by Henryt999 View Post
    How do you find the maximum value of (ln(x))/x^2

    First i derived: That gave me: (1-2ln(x))/x^3
    Now i figure x' is zero at: x = 0
    or ln(x) = 0,5 = e^0,5

    But the correct answer is suppose to be 1/(2e)

    What am i doing wrong?
    With Kind regards Henry from Sweden
    f'(x) = 0 at \ln{x} = \frac{1}{2}

    x = e^{\frac{1}{2}}

    max value is ...

    f\left(e^{\frac{1}{2}}\right) = \frac{\ln\left(e^{\frac{1}{2}}\right)}{e} = \frac{\frac{1}{2}}{e} = \frac{1}{2e}
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    Quote Originally Posted by Raoh View Post
    there is a global maximum at \sqrt{e}.
    \sqrt{e} is the root of your first derivative.
    and since f' changes its sign at \sqrt{e} .
    f(\sqrt{e}) is your maximum.
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  6. #6
    Moo
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    Quote Originally Posted by Raoh View Post
    and since f' changes its sign at img.top {vertical-align:15%;} \sqrt{e} " alt=" \sqrt{e} " />.
    f(\sqrt{e}) is your maximum.
    It could also be a minimum.
    \sqrt{e} is a maximum if the second derivative at this point is negative.
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  7. #7
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    Quote Originally Posted by Moo View Post
    It could also be a minimum.
    \sqrt{e} is a maximum if the second derivative at this point is negative.
    f''(\sqrt{e})<0
    it's a global maximum.
    thanks.
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