Results 1 to 7 of 7

Thread: Max Value of: Ln(x)/(x^2)

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    180

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

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

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

    Now what is $\displaystyle f(x_{\max})$, where $\displaystyle f(x)=\frac{\ln(x)}{x^2}$ ? 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 $\displaystyle f''(x_{\max})<0$
    Last edited by Moo; Dec 29th 2009 at 07:23 AM. Reason: you're -> your
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    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 $\displaystyle \sqrt{e}$.
    $\displaystyle \sqrt{e}$ is the root of your first derivative.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    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
    $\displaystyle f'(x) = 0$ at $\displaystyle \ln{x} = \frac{1}{2}$

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

    max value is ...

    $\displaystyle f\left(e^{\frac{1}{2}}\right) = \frac{\ln\left(e^{\frac{1}{2}}\right)}{e} = \frac{\frac{1}{2}}{e} = \frac{1}{2e}$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by Raoh View Post
    there is a global maximum at $\displaystyle \sqrt{e}$.
    $\displaystyle \sqrt{e}$ is the root of your first derivative.
    and since $\displaystyle f'$ changes its sign at $\displaystyle \sqrt{e}$ .
    $\displaystyle f(\sqrt{e})$ is your maximum.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Raoh View Post
    and since f' changes its sign at $\displaystyle $\displaystyle \sqrt{e}$ $.
    $\displaystyle f(\sqrt{e})$ is your maximum.
    It could also be a minimum.
    $\displaystyle \sqrt{e}$ is a maximum if the second derivative at this point is negative.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by Moo View Post
    It could also be a minimum.
    $\displaystyle \sqrt{e}$ is a maximum if the second derivative at this point is negative.
    $\displaystyle f''(\sqrt{e})<0$
    it's a global maximum.
    thanks.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum