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

A central cross-section of the sphere gives the constraint $r^2+h^2/4=R^2$, and the cylinder's volume is $V=\pi r^{2}h$. At an interior optimum with $r,h>0$, the Lagrange multiplier equations for this constrained optimization problem areThe first gives $\lambda =\pi h$; substitution into the second gives $r^2=h^2/2$. The constraint then givesThe feasible nonnegative $(r,h)$ form a compact arc, whose endpoints have zero volume. Since this is the only positive stationary candidate, it is the global maximum. Its volume and fraction of the sphere's volume are

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 $4\sqrt{2}$.