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
- $f_x'(x,y)=3x^2y^3$
- $f_y'(x,y)=3y^2x^3$
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?
Best Answer
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 $(0,0)$