Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 17 / a

The zero set consists of the three lines $x=0$, $y=0$ and $y=x-1$. Their pairwise intersections areThe bounded triangle has $0\le x\le 1$ and $x-1\le y\le 0$. Its edges and interior contours are shown below.

The gradient is $\nabla f=(y(1-2x+y),x(1-x+2y))$. At the three specified points it is respectivelyEach vector is perpendicular to its straight boundary edge and points out of the triangle, where $f$ increases from its negative interior values. The arrows in the preceding figure mark these directions.

Inside the triangle, both $x$ and $y$ are nonzero. Thus $\nabla f=0$ reduces to $1-2x+y=0$ and $1-x+2y=0$, whose unique solution is $(1/3,-1/3)$. At that pointThe Hessian matrix has positive eigenvalues $1$ and $1/3$, so this is a strict local minimum. It is also the minimum on the closed triangle: $f=0$ on the boundary, $f<0$ inside, and there is no other interior stationary point. Therefore

The interior values satisfy $-1/27<f<0$ away from the minimum. The contours are closed curves surrounding $(1/3,-1/3)$, approximately ellipses near that point, and approach the three edges as the level tends to zero. To see their shape globally, put $u=x$, $v=-y$, $w=1-x+y$. These are positive inside and satisfy $u+v+w=1$, while $f=-uvw$. Since $\log (uvw)=\log u+\log v+\log w$ is strictly concave, its nonempty superlevel sets are convex and surround the unique maximum of $uvw$. Thus the negative contours nest without additional components, as in the figure.