Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability (resit) / 6 / a

The critical points satisfy both first partial derivatives equal to zero:Substitute $x=-y/2$ into the second equation to get $y(3y-1/2)=0$. Thus the candidates are $(0,0)$ and $(-1/12,1/6)$.

The Hessian matrix and its determinant areAt $(0,0)$ the determinant is $-1$, so the Hessian matrix is indefinite and the point is a saddle point. At $(-1/12,1/6)$ the determinant is $1$ and $h_{xx}=2>0$, so the Hessian matrix is positive definite and the point is a strict local minimum, with valueThere is no local maximum. Along $x=0$, $h(0,y)=y^3$ is unbounded above and below, so there are no global extrema either.