Q.Find two positive numbers whose sum is $16$ and the sum of whose cubes is minimum.

Express the cone's base-radius in terms of its height using the sphere's geometry, write volume as a function of height alone, and maximize.

(Answering the general-students version.) Let the sphere have radius $r$ and centre $O$, and let the cone have height $h$ and base radius $x$. If the cone's apex and the centre are positioned so the base is at perpendicular distance $(h-r)$ from the centre, then by the Pythagorean relation on the base circle:

$x^2=r^2-(h-r)^2=2rh-h^2,\qquad 0<h<2r$

Volume: $V=\dfrac13\pi x^2h=\dfrac13\pi(2rh-h^2)h=\dfrac13\pi(2rh^2-h^3)$.

…

Express curved surface area in terms of $r$ alone (using the fixed-volume constraint to eliminate $h$), then minimize.

Volume $V=\dfrac13\pi r^2h$ (fixed) $\Rightarrow h=\dfrac{3V}{\pi r^2}$.

Curved surface area $S=\pi rl=\pi r\sqrt{r^2+h^2}$. Work with $S^2=\pi^2r^2(r^2+h^2)=\pi^2r^4+\pi^2r^2h^2$.

Substituting $h^2=\dfrac{9V^2}{\pi^2r^4}$: $S^2=\pi^2r^4+\dfrac{9V^2}{r^2}$. Let $f(r)=\pi^2r^4+\dfrac{9V^2}{r^2}$ (minimizing $S$ is equivalent to minimizing $f$ since $S>0$).

$f'(r)=4\pi^2r^3-\dfrac{18V^2}{r^3}=0 \Rightarrow 4\pi^2r^6=18V^2 \Rightarrow r^6=\dfrac{9V^2}{2\pi^2}$.

…

Express the cone's volume as a function of the semi-vertical angle $\theta$ alone (slant height $l$ fixed), then maximise using calculus.

For slant height $l$ and semi-vertical angle $\theta$: radius $r=l\sin\theta$, height $h=l\cos\theta$.

$V=\frac13\pi r^2h=\frac13\pi(l\sin\theta)^2(l\cos\theta)=\frac{\pi l^3}{3}\sin^2\theta\cos\theta$.

Differentiate w.r.t. $\theta$ (product rule):

$\frac{dV}{d\theta}=\frac{\pi l^3}{3}\left[2\sin\theta\cos\theta\cdot\cos\theta+\sin^2\theta\cdot(-\sin\theta)\right]=\frac{\pi l^3}{3}\sin\theta\left(2\cos^2\theta-\sin^2\theta\right)$.

…

Express volume in terms of height using $r^2 = R^2 - \tfrac{h^2}{4}$, maximise via $\tfrac{dV}{dh}=0$.

For a cylinder of radius $r$, height $h$ inscribed in a sphere of radius $R$: $r^2 + \left(\tfrac h2\right)^2 = R^2 \Rightarrow r^2 = R^2 - \tfrac{h^2}{4}.$

$V = \pi r^2 h = \pi\left(R^2 - \tfrac{h^2}{4}\right)h = \pi R^2 h - \tfrac{\pi h^3}{4}.$

$\dfrac{dV}{dh} = \pi R^2 - \tfrac{3\pi h^2}{4} = 0 \Rightarrow h^2 = \dfrac{4R^2}{3} \Rightarrow h = \dfrac{2R}{\sqrt3}.$

$\dfrac{d^2V}{dh^2} = -\tfrac{3\pi h}{2} < 0$, confirming a maximum.

…