Past exam of the mathematical methods II course of the University of Oxford / 2017 resit / 6 / Solution

The gradient equations are $f_x=2x(1-y)=0$ and $f_y=4y-x^2=0$. The critical points are $(0,0)$ and $(\pm 2,1)$. Their Hessian matrix isAt the origin $H=\mathrm{diag}(2,4)$ is positive definite. At either other point its determinant is $-16$, so its eigenvalues have opposite signs. ThusThe minimum value is zero and both saddle values are two. The origin is not a global minimum: at any fixed $y>1$, the term $(1-y)x^2$ makes $f$ arbitrarily negative as $|x|$ increases.

For the contour plot, the local contours around the origin are approximately $x^2+2y^2=c$. The exact saddle contour factors asso it is the union of $y=1$ and $y=x^2/2-1$, crossing at the two saddles. Other levels obey $x^2=(c-2y^2)/(1-y)$ where $y\ne 1$; the right panel of the original figure in part 3(a) shows how these branches join around the minimum and pass through the saddle contour.