 Solution

The constraint is a compact set, so the continuous objective attains both extrema by the extreme value theorem. Use Lagrange multipliers with $q(x,y)=9x^2+y^2$:Since the second component makes $\lambda \ne 0$, the two equations give $4/(18x)=1/(2y)$, or $y=9x/4$. Substitution into the constraint yieldswith matching signs. The objective values are thereforeThe constraint gradient never vanishes on the ellipse, so these are all the possible extrema. As a direct check, set $u=3x$, $v=y$; then $u^2+v^2=36$ and the Cauchy-Schwarz inequality gives $|4u/3+v|\le 6\sqrt{16/9+1}=10$, with equality at the same points.