Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 8A / b / Solution

The gradient components areIf $x=0$ or $y=0$, their simultaneous zeros give $(0,0),(0,1),(1,0)$. If $xy\ne 0$, solving $2x+y=1$, $x+2y=1$ gives $(1/3,1/3)$. These are all the critical points. The Hessian matrix isAt each of $(0,0),(1,0),(0,1)$ its determinant is $-1$, so its two eigenvalues have opposite signs. All three are saddle points of a scalar function. At $(1/3,1/3)$,This point is a strict local maximum, of value $\boxed{1/27}$.

The zero level set is exactlyThese three lines form a triangle containing the maximum. Inside it $f>0$; near the maximum positive level sets are nested closed curves, with their leading shapes elliptical because the Hessian matrix is negative definite. At each vertex the zero lines cross, separating alternating signs, and nearby nonzero level sets have hyperbolic branches. For example, the leading quadratic terms are $xy$ at $(0,0)$, $-\eta (\xi +\eta )$ for $\xi =x-1,\eta =y$ at $(1,0)$, and $-\xi (\xi +\eta )$ for $\xi =x,\eta =y-1$ at $(0,1)$.