# Proving minima, maxima, and saddle points don't exist.

I've got the next function: $f(x,y)=x^3y^3$ where $x,y\in \mathbb R$

I need to determine whether there is minima, maxima or saddle point.

Easily enough, after doing the partial derivatives

1. $f_x'(x,y)=3x^2y^3$
2. $f_y'(x,y)=3y^2x^3$

I get the point $(0,0)$.

Now, how can i "officially" prove that that point isn't a minima, maxima or saddle point? Is it enough to show that hessian matrix gives 0? Do i need to show that the function can get higher [/lower] values using other points on the function?

Use the definition of maxima/minima. Try to find an open neighborhood $U$ containing $(0,0)$ s.t. for all $(x,y)\in U$ you have either

$$f(x,y)-f(0,0)>0$$

or

$$f(x,y)-f(0,0)<0$$

I supposed that the domain of $f$ is the whole $\mathbb R^2$. The answer to your question is then equivalent to study either

$$x^3y^3>0$$

or

$$x^3y^3<0$$

With an $(x,y)$ plot and focusing on what happens around $(0,0)$ you can arrive quite easily to the answer (=negative one, i.e. there exists no such neighborhood $U$, so $(0,0)$ cannot be either a max or a min).

Edit: I add the plot of $f$

Wolfram plot and contour plot of $f$ around $(0,0)$

• Thanks. But what about proving there's no saddle point? – StationaryTraveller Jun 20 '13 at 15:19
• $(0,0)$ is a saddle point because it is a stationary point of $f$ but no maximum/minimum. Please check en.wikipedia.org/wiki/Saddle_point – Avitus Jun 20 '13 at 15:22
• And something i want to clarify, By "Open neighborhood" you mean an some-kind of intervals for x,y? Then i need to show that two or more 'intervals' give different results? – StationaryTraveller Jun 20 '13 at 15:25
• Open neighborhood in the metric topology of $\mathbb R^2$, i.e. a union of open "balls". Here you can consider any open ball with center in $(0,0)$ and radius $R>0$ to arrive at the thesis – Avitus Jun 20 '13 at 15:29
• Please, check the Wiki article in my comment above. If you have a stationary point of a given function which is neither a maximum nor a minimum, then it is called saddle point. In your case you proved that $(0,0)$ is a stationary point, i.e. $f_x(0,0)=f_y(0,0)=0$, but the discussion in my answer allows you to conclude that $(0,0)$ cannot be neither a max nor a min. – Avitus Jun 20 '13 at 15:33