= Solution
The constraint is a <compact set>, so the continuous objective attains both extrema by the <extreme value theorem>. Use <Lagrange multipliers> with $q(x,y)=9x^2+y^2$:
$$
(4,1)=\lambda(18x,2y).
$$
Since the second component makes $\lambda\ne0$, the two equations give $4/(18x)=1/(2y)$, or $y=9x/4$. Substitution into the constraint yields
$$
\frac{225}{16}x^2=36,\qquad x=\pm\frac85,\quad y=\pm\frac{18}5,
$$
with matching signs. The objective values are therefore
$$
\boxed{\max(4x+y)=10\text{ at }(8/5,18/5),\qquad
\min(4x+y)=-10\text{ at }(-8/5,-18/5)}.
$$
The 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 $|4u/3+v|\leq6\sqrt{16/9+1}=10$, with equality at the same points.
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