= Solution
Use the <Lagrange multiplier> condition $\nabla f=\lambda\nabla g$ for $g=4x^2+y^2/16$:
$$
y=8\lambda x,\qquad x=\lambda y/8,\qquad4x^2+y^2/16=1.
$$
Neither coordinate can vanish at a constrained <stationary point>: either would force both to vanish, contradicting the constraint. Substitution therefore gives $\lambda^2=1$, so $y=8x$ or $y=-8x$. In either case $8x^2=1$, giving
$$
\boxed{(x,y)=\left(\frac{\varepsilon}{2\sqrt2},\ 2\sqrt2\,\delta\right),\qquad
\varepsilon,\delta\in\{-1,1\},\qquad f=\varepsilon\delta}.
$$
Equal signs give the two points attaining the <global maximum> $f=1$; opposite signs give the two <global minima> with $f=-1$. These exhaust the extrema because the <ellipse> is compact, its constraint <gradient> never vanishes on it, and $f$ is smooth. The bound can also be checked by setting $X=2x$, $Y=y/4$: then $X^2+Y^2=1$ and $f=2XY$, so $|f|\leq1$.
Back to article page
