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

1. ## 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

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

3. 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.

4. 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 ...

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

5. 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.

6. Originally Posted by Raoh
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.

7. 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.