# Find extremes of $f(x,y,z) = x^4 +y^4 -2x^2 +2x^2y^2 +z^2$

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

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

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?

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.

• but let say that we have $\lambda_1 = 0 \lambda_2 = 5 \lambda_3 = 10$ - then we have minimum or not? due to $\lambda_1 = 0$ – MrQuestion1999 Jun 12 at 21:23
• 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 ;) – Sam Skywalker Jun 12 at 21:36
• 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 ^^ – Sam Skywalker Jun 12 at 21:38

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.

• but let say that we have $\lambda_1 = 0 \lambda_2 = 5 \lambda_3 = 10$ - then we have minimum or not? due to $\lambda_1 = 0$ – MrQuestion1999 Jun 12 at 21:22
• 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. – José Carlos Santos Jun 12 at 22:10