# Thread: help finding the Maxima

1. ## help finding the Maxima

I would be greatful for any help with the following problem. I know i need to find the dirivative of v and then from this i can transpose to find the unknown value of x. Just don't seem to be able!

Q. The velocity, v, of a signal in a cable at a distance x is given by :

v = K x ln (1/x) where 0<x<1 ( x is the algibraic x )

where K is a positive constant. Find the value of x which gives the maximum velocity.

Thanks again, James.

2. NOT trying to bump, Just wanted to give it in the proper way.

$V = K x ln( \frac{1}{x} )$

3. Originally Posted by james jarvis
NOT trying to bump, Just wanted to give it in the proper way.

$V = K x ln( \frac{1}{x} )$
You'll see that in every reply you post there will be a button that says "Edit", right next to "Quote"

4. Originally Posted by james jarvis
NOT trying to bump, Just wanted to give it in the proper way.

$V = K x ln( \frac{1}{x} )$
$\frac{dV}{dx} = K~ ln \left ( \frac{1}{x} \right ) + Kx \cdot \frac{1}{\frac{1}{x}} \cdot -\frac{1}{x^2}$

$\frac{dV}{dx} = K~ ln \left ( \frac{1}{x} \right ) - K$

Set this equal to 0:
$K~ ln \left ( \frac{1}{x} \right ) - K = 0$

$ln \left ( \frac{1}{x} \right ) - 1 = 0$

$ln \left ( \frac{1}{x} \right ) = 1$

$\frac{1}{x} = e$

$x = \frac{1}{e}$

Is this a relative max or a relative min? I leave the answer to that question up to you.

-Dan