Semidefinite Hessian Matrix, Global and Local Extrema

Question

Given the function $f(x,y) := x^2y$, which is defined on all $\{(x,y):x^2+y^2\le1\}$, the first partial derivatives are $0$ where $x=0$, however, the Hessian at all these points has the Eigenvalue $0$, it is semidefinite. We are supposed to find all local and global extrema. How does one handle the points where the Hessian is semidefinite? Do they have a name? Is it simply impossible to make a statement on local maxima/minima inside the compact set?

Answer 1

We have $f(t,t)=t^3$ and $f(0,t)=0$.

If $t>0$ we get $f(t,t)>f(0,t)$ and if $t<0$ we get $f(t,t)<f(0,t)$.

Conclusion: in $(0,t)$ the function $f$ has no local extremum

Answer 2

I think you must search the optimum on the vurve $$x^2+y^2=1$$ so you will get $$f(\pm\sqrt{1-y^2},y)=(1-y^2)y$$