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I am preparing for an exam on Tuesday and I'm having problems with this excercise:

Decide, whether there are extremes for the function $f:\mathbb{R^3} \rightarrow \mathbb{R}, f(x,y,z) = xyz \hspace{2mm}$ on an elipsoid $3x^2 + 3y^2 + z^2 = 1 \hspace{2mm}$ If there are any, find them.

My solution so far:

$3x^2 + 3y^2 + z^2 -1 = 0$
$L(x,y,z, \lambda) = xyz - \lambda(3x^2 + 3y^2 + z^2 -1)$

$f'_x = yz \hspace{21mm} N'_x =6x$
$f'_y = xz \hspace{21mm} N'_y =6y$
$f'_z = xy \hspace{21mm} N'_z =2z$

$(yz, xz, xy) = k(6x, 6y, 2z)$

$yz = 6kx$
$xz = 6ky$
$xy = 2kz$
$3x^2 + 3y^2 + z^2 = 1$

And I have a really hard time getting the points out of these four equations - I always run into a dead end. Can someone please give me a hint? Also I'm sorry for the formatting, but I haven't used LateX in a while.

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up vote 2 down vote accepted

It's hard to give general advice for solving systems of equations. Usually one looks to one by one eliminate variables and substitute them into other equations. However, that doesn't work very well for this system. A second general trick is to try to "symmetrize" equations -- that is -- make them or one side of them look the same.

Consider multiplying your first equation by $x$, the next by $y$, and the third by $z$. You'll get

$xyz=6kx^2$, $xyz=6ky^2$, and $xyz=2kz^2$. Now dividing each by $2k$ you get that $3x^2=3y^2=z^2$. At this point you can plug them into the constraint equation and find the answer.

$3x^2+3y^2+z^2=1$ becomes $z^2+z^2+z^2=1$ so that $z^2=1/3$ and so $z=\pm 1/\sqrt{3}$. Next $3x^2=z^2=1/3$ etc.

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First of all - THANKS. Now as I look at it the solution is really easy. I kept trying to substitute for z and then solving it, but it got really long and messy. This is quite simple on the other hand. I would remember it. Really - thanks! – Alexandar Živkovič Oct 20 '11 at 23:36

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