Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 3 / c / Solution

Write $A=a^2>0$ and $E=e^{-(x^2+y^2)/A}>0$. The partial derivatives factor asA critical point has both partial derivatives zero. If $x,y\ne 0$, this would require $x^2-2y^2=A$ and $x^2-2y^2=-2A$, which is impossible. On the axes the remaining conditions giveAt either critical point with $x\ne 0$, the Hessian matrix isIt is negative definite, so the second-derivative test shows that both $(\pm |a|,0)$ are strict local maxima, of value $a^2/e$. They also attain the global maximum: $f\le x^2{e}^{-x^2/A}\le A/e$. For completeness, the Hessian matrix at $(0,\pm |a|)$ is $\mathrm{diag}(6/e,8/e)$, giving strict local minima, while at the origin it is $\mathrm{diag}(2,-4)$, giving a saddle point.