Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 13 / a

Put $X=x-1$ and $H=e^{-(X^2+y^2)/2}>0$. The gradient components are $z_x=y(1-X^2)H$ and $z_y=X(1-y^2)H$. If either $X$ or $y$ vanishes, both must vanish. Otherwise both squared coordinates must be one. Therefore the five stationary points in the original $(x,y)$ coordinates are

At $(X,y)=(0,0)$ the Hessian matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(x,y)=(1,0)$ is a saddle point. At the other four points the mixed second derivative vanishes and the two diagonal entries both equal $-2Xy/e$. HenceThe four extrema are global: $|z|=(|X|e^{-X^2/2})(|y|e^{-y^2/2})\le e^{-1}$ because each factor has maximum $e^{-1/2}$ at absolute argument one.

The zero level set consists of the lines $x=1$ and $y=0$. The height is positive where $(x-1)y>0$ and negative where that product is negative. In each of the four resulting regions, closed nonzero contours surround its extremum; levels closer to zero extend further from the extremum towards the zero-level axes and towards large distance. The height tends to zero at infinity. The figure marks the four extrema and central saddle point, and includes the downhill path used below.