Minimise the cone circumscribed about a sphere

Let $V$ be the cone apex, $C$ its base centre, and $T$ the point where the sphere touches a sloping generator in an axial section. Then $VO=h-a$ and $OT=a$, with $OT\perp VT$, so $VT=\sqrt{(h-a)^2-a^2}=\sqrt{h^2-2ah}$. The similar right triangles $VTO$ and $VCB$ (where $B$ is a base-rim point) give $VT/h=a/r$. Squaring,

$\frac{h^2-2ah}{h^2}=\frac{a^2}{r^2},\qquad\frac{h-2a}{h}=\frac{a^2}{r^2},\qquad r^2=\frac{a^2h}{h-2a}.$

The physical domain is $a>0$, $h>2a$. Merely requiring $h>0$ would allow impossible cones.

Substitute the relation into the cone volume:

$V=\frac{\pi a^2}{3}\frac{h^2}{h-2a},\qquad \frac{dV}{dh}=\frac{\pi a^2}{3}\frac{2h(h-2a)-h^2}{(h-2a)^2}=\frac{\pi a^2h(h-4a)}{3(h-2a)^2}.$

The algebraic roots are $h=0,4a$; reject $0$ as outside the physical domain. Thus the only admissible stationary height is $h=4a$.

For $h>2a$, the factor $\pi a^2h/[3(h-2a)^2]$ is positive, so $dV/dh$ has the sign of $h-4a$. The source sign chart is negative–zero–positive across $4a$: the volume falls on $(2a,4a)$ and rises on $(4a,\infty)$. This proves a minimum throughout the feasible domain.

The quotient rule gives

$\begin{aligned}&\frac{d^2V}{dh^2}\\ & =\frac{\pi a^2}{3}\frac{(h-2a)^2(2h-4a)-2h(h-4a)(h-2a)}{(h-2a)^4}\\ & =\frac{8\pi a^4}{3(h-2a)^3}.\end{aligned}$

At $h=4a$, this is $\pi a/3>0$. Substituting into the volume gives $V_{\min}=\tfrac{\pi a^2}{3}(4a)^2/(4a-2a)=\tfrac83\pi a^3$. Both printed minimum tests are retained.