Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 3 / b

The critical points satisfySubstitute $x=-y/2$ to obtain $y(3y-1/2)=0$. Thus the only critical points are $(0,0)$ and $(-1/12,1/6)$.

The Hessian matrix and Hessian determinant areAt $(0,0)$, $D=-1$, so this is a saddle point. At $(-1/12,1/6)$, $D=1$ and $f_{xx}=2>0$, so this is a strict local minimum. ThereforeThe value at the local minimum is $-1/432$; there is no local maximum.