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I was trying to find the maximum, the minimum and the saddle points of the following function:

$f(x,y,z)=x^3 + y^3 + z^3 -9xy -9xz +27x$

I derivate the function, so:

$f_x=3x^2-9y-9z+27$
$f_y=3y^2-9x$
$f_z=3z^2-9x$

So in order to find these point I need to check when $f_x=0$;$f_y=0$ and $f_z=0$; but the only thing I've been able to prove is that y=z by suming the last two but when I tried to use it in the first one I get : $y=\frac{x^2}{6}+\frac{3}{2}$ which does not have any root. I've tried other ways but I always ended up in the same result. And honestly I don't know what to anymore. I would really appreciate any advice you can give me

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You can reduce $f_x, f_y, f_z$ by dividing out the $3$ to:

$$x^2-3y-3z+9 = 0 \\ y^2-3x = 0 \\z^2-3x =0$$

Update

From $z = \pm y$, we substitute in the first equation and have:

$$x^2 - 3 y+ 3 (\pm~ y) + 9 = 0$$

This gives two cases to check:

$$x^2 - 6y + 9 = 0, x^2 + 9 = 0$$

The second does not work, for the first, we know that $x = \dfrac{1}{3} y^2$. Substitute and you get $y = 3, 1.63107$ and two imaginary roots that we can immediately ignore (real numbers only). Now use those two $y$ values, substitute to find the corresponding $x$ and $z$ values.

One of those has a solution $x = 0.886793$ and another solution $x = 3$. Find the corresponding $z$ values. All other choices/approaches lead to these roots and other imaginary roots that we can ignore.

So, we have the following roots:

$$(x, y, z) = (3,3,3) ~~\mbox{or} ~~ (0.886793, 1.63107, 1.63107)$$

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  • $\begingroup$ Thank you so much for your help, may I ask you something, I was trying to follow your solution but how can you from $z=y or z=-y$ get to $x=y=z=3$? $\endgroup$ – Felipe Nov 13 '16 at 18:59
  • $\begingroup$ by the way, how can you know that all the point you find are the only ones? I mean is it possible to be more points? $\endgroup$ – Felipe Nov 13 '16 at 19:00
  • $\begingroup$ See update - even easier using this approach. $\endgroup$ – Moo Nov 13 '16 at 20:56

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