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Find extremes (and decide if it is minimum or maximum) of $f(x,y,z) = x^4 +y^4 -2x^2 +2x^2y^2 +z^2$

I made hessian matrix

$$ \left( \begin{array}{ccc} 12 x^2+y^2-1 & 8 x y & 0 \\ 8 x y & 4 x^2 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$$

And for each potential value of extreme I calculated three another matrices
for $0,0,0$ $$\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ for $1,0,0$ $$\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

for $-1,0,0$ $$ \left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}\right) $$

Problem

What if for example in second case I got own values: $ \lambda_1 = 0 \wedge \lambda_2 = 11 \wedge\lambda_3 = 4 $ - is there minimum or there is no extreme?

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Your Hessian matrix contains a computation mistake. It actually is $$\left( \begin{array}{ccc} 12 x^2+4y^2-4 & 8 x y & 0 \\ 8 x y & 12y^2 + 4x^2 & 0 \\ 0 & 0 & 2 \\ \end{array} \right), $$

which yields a different behaviour at the critical points. At $(0,0,0)$ it is still singular, since it is $$\left( \begin{array}{ccc} -4 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \\ \end{array} \right), $$ and you will need to assess whether it is an extreme or not by hand (this looks to me as a degenerate saddle point, so we should be able to find a path through the origin along which $f$ grows and other along which $f$ decreases).

However, at the other points we have that $$H_{(1,0,0)} (f)\left( \begin{array}{ccc} 8 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 2 \\ \end{array} \right), $$ which is a positive definite form; thus, $f$ has a minimum at $(1,0,0)$.

Notice that $H_{(-1,0,0)}(f) = H_{(1,0,0)}(f)$, so $f$ has a minimum at $(-1,0,0)$ as well.

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  • $\begingroup$ but let say that we have $\lambda_1 = 0 \lambda_2 = 5 \lambda_3 = 10$ - then we have minimum or not? due to $\lambda_1 = 0 $ $\endgroup$ – MrQuestion1999 Jun 12 at 21:23
  • $\begingroup$ I suspect it is just a degenerate critical point. Not a minimum; not a maximum either. It is as if $f$ were flat in the $x$ direction while decreasing in the $y$ direction, according to Wolfram Alpha (I left out the $z$ coordinate since it adds nothing to the question of $(0,0,0)$ as a candidate to minimum. Try to find a path through the origin for which it is not a minimum ;) $\endgroup$ – Sam Skywalker Jun 12 at 21:36
  • $\begingroup$ Hint: the path can be a straight line. Find two straight lines such that it is a minimum for one but not for the other. Good luck ^^ $\endgroup$ – Sam Skywalker Jun 12 at 21:38
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Both cases can occur. If $f(x,y,z)=x^4+\frac{11}2y^2+2z^2$ you have a minimum at $(0,0,0)$, and if $f(x,y,z)=-x^4+\frac{11}2y^2+2z^2$ then $(0,0,0)$ is a saddle point.

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  • $\begingroup$ but let say that we have $\lambda_1 = 0 \lambda_2 = 5 \lambda_3 = 10$ - then we have minimum or not? due to $\lambda_1 = 0 $ $\endgroup$ – MrQuestion1999 Jun 12 at 21:22
  • $\begingroup$ I gave you an example in my answer of a situation in which we have a minimum and a situation in which we don't. Or, as I wrote, both cases can occur. $\endgroup$ – José Carlos Santos Jun 12 at 22:10

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