Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 19 / b

Use the Lagrange multiplier condition $\nabla f=\lambda \nabla g$ for $g=4x^2+y^2/16$: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 $8x^2=1$, givingEqual 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 $|f|\le 1$.

For a point in the first quadrant, $xy$ is the area of the axis-aligned rectangle with opposite vertices at the origin and $(x,y)$. Thus the maximum says that the largest such quarter-rectangle with its outer corner on the ellipse has area one.

Equivalently, reflect this rectangle in both axes. Its four vertices $(\pm |x|,\pm |y|)$ lie on the ellipse, and its full area is $4|xy|$. ThereforeIts width is $1/\sqrt{2}$ and its height is $4\sqrt{2}$.