Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 4 / d

Let $q=x^2+4y^2$. The objective has the exact sum of squares decompositionFor every feasible point, $q\le 8$, so this is at least $3-8/2=-1$. Equality requires simultaneously $q=8$ and $x+2y=0$. Substitution gives $y^2=1$ and $x=-2y$. Both resulting points are feasible. Therefore the global minimum is $-1$, attained exactly at $(2,-1)$ and $(-2,1)$. The lower bound and its equality conditions prove global attainment of the minimum, rather than merely identifying stationary points on the boundary.