# maximum values

• Sep 28th 2010, 04:57 AM
ulysses123
maximum values
I am trying to find the set of points for which the maximum value of the curvature

$K=\fract{3} { (x^{-2}+y^{-2}+z^{-2})^{-2} }$

occurs for the function $xyz=1$

I have equated the gradients and subject to the restriction of the function xyz=1 i only get one solution the point (1,1,1), since when i solve the equations i get $x^3=y^3=z^3$

but i am not sure if i have done this correctly.Anyway any help would be appreciated.
• Sep 28th 2010, 07:01 AM
CaptainBlack
Quote:

Originally Posted by ulysses123
I am trying to find the set of points for which the maximum value of the curvature

$K=\fract{3} { (x^{-2}+y^{-2}+z^{-2})^{-2} }$

occurs for the function $xyz=1$

I have equated the gradients and subject to the restriction of the function xyz=1 i only get one solution the point (1,1,1), since when i solve the equations i get $x^3=y^3=z^3$

but i am not sure if i have done this correctly.Anyway any help would be appreciated.

There is something wrong with your LaTeX you have a \frac{}{} which does not render.

Other than that the AM-GM inequality will (probably - since the question has some ambiguity due to the LaTeX problem) give your result.

CB
• Sep 28th 2010, 07:48 PM
ulysses123
Yes i noticed that, but since this is the correct equation i didn't alter it.
• Sep 29th 2010, 12:33 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
There is something wrong with your LaTeX you have a \frac{}{} which does not render.

Other than that the AM-GM inequality will (probably - since the question has some ambiguity due to the LaTeX problem) give your result.

CB

The maximum of $K$ occurs when: $x^{-2}+y^{-2}+z^{-2}$ is at a minimum.

The AM-GM inequality tells us that:

$x^{-2}+y^{-2}+z^{-2}\ge 3 \root 3 \of {x^{-2}y^{-2}z^{-2}} =3$

with equality only when $x^{-2}=y^{-2}=z^{-2}$ , or $x=y=z=1$.

So the maximum of $Z$ occurs at $x=y=z=1$

CB