Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 12

Let $D=1+x^2+y^2$. The gradient has components $f_x=[D-2x(x+y)]/D^2$ and $f_y=[D-2y(x+y)]/D^2$. At a critical point, subtraction gives $(x-y)(x+y)=0$. The case $x+y=0$ cannot make either numerator zero, so $x=y=t$. The remaining equation is $1+2t^2=4t^2$, givingThese are the only stationary points; the denominator is positive everywhere.

At $(t,t)$ with $t^2=1/2$, differentiating the gradient yields the Hessian matrix $H=-tI$. At the positive point it is negative definite, and at the negative point it is positive-definite; in both cases $f_{xy}=0$ and $\det H=1/2$. Therefore the positive point is a maximum, the negative point is a minimum, and there are no saddle points.

They are global extrema: with $r=\sqrt{x^2+y^2}$, the Cauchy-Schwarz inequality gives $|f|\le \sqrt{2}r/(1+r^2)\le 1/\sqrt{2}$, with equality at exactly those two points. This is the extremum bound underlying the level sets of a linear numerator over a radial quadratic.

For a nonzero level $h$, completing the square givesThus the level sets of a linear numerator over a radial quadratic are circles centred on $y=x$, real only for $0<|h|\le 1/\sqrt{2}$. The extreme levels shrink to points. The zero contour is the straight line $x+y=0$, separating positive and negative heights; near it, small nonzero levels are large circles. Heights tend to zero at infinity.