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

**Henryt999** · December 29th 2009, 06:47 AM

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

**Moo** · December 29th 2009, 06:54 AM

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 $f''(x_{\max})<0$

**Raoh** · December 29th 2009, 06:58 AM

Originally Posted by Henryt999

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.

**skeeter** · December 29th 2009, 07:02 AM

Originally Posted by Henryt999

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 ...

$f\left(e^{\frac{1}{2}}\right) = \frac{\ln\left(e^{\frac{1}{2}}\right)}{e} = \frac{\frac{1}{2}}{e} = \frac{1}{2e}$

**Raoh** · December 29th 2009, 07:08 AM

Originally Posted by Raoh

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.

**Moo** · December 29th 2009, 07:09 AM

Originally Posted by Raoh

and since f' changes its sign at $<html> <head> <style type=$ 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.

**Raoh** · December 29th 2009, 07:16 AM

Originally Posted by Moo

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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