Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 5 / f

The continuous function attains a global maximum and minimum on the closed disc by the extreme value theorem. All four unconstrained stationary points have squared distance eight, so none lies in its interior. The extrema must therefore lie on $x^2+y^2=6$.

The constraint gradient is nonzero on that circle, so Lagrange multipliers apply:Subtracting gives $(x-y)((x+y)/4-2\lambda )=0$. If $x=y$, the constraint gives $x=y=\pm \sqrt{3}$. Otherwise $2\lambda =(x+y)/4$; substitution into either equation gives $xy=-4$. This is impossible, since $|xy|\le (x^2+y^2)/2=3$ on the circle. Thus the diagonal points are the only candidates. Comparing their values gives