# Semidefinite Hessian Matrix, Global and Local Extrema

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?

• Another thing to keep in mind: global and local maxima and minima might occur on the boundary of your constraint set, and thus not be found by the usual calculus recipes. – kimchi lover Jun 6 at 13:15
• @kimchi lover Thanks for the hint, but should I include or exclude the (0,y) values from my solution? – Ruben Kruepper Jun 6 at 13:19
• Your title, question, and comment are inconsistent. Are you asked for local/global extrema, or critical points, or both? Critical points, at least, you can find by calculus. – kimchi lover Jun 6 at 13:23
• @kimchi lover Apologies, English is not the language I'm studying in. – Ruben Kruepper Jun 6 at 13:25

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 $$f(t,t).
Conclusion: in $$(0,t)$$ the function $$f$$ has no local extremum
I think you must search the optimum on the vurve $$x^2+y^2=1$$ so you will get $$f(\pm\sqrt{1-y^2},y)=(1-y^2)y$$
• The global extrema are situated on the curve $$x^2+y^2=1$$ – Dr. Sonnhard Graubner Jun 6 at 13:43