Problem Statement
Show that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.
Verified Solution & Marking Scheme
Geometry of inscribed cone and sphere relation
Let $R$ be the radius of the sphere. Let $r$ and $h$ be the radius of the base and height of the cone respectively.
From the right triangle formed by the center of the sphere, the center of the cone's base, and a point on the rim:
$$r^2 + (h - R)^2 = R^2 \implies r^2 = R^2 - (h - R)^2 = 2Rh - h^2$$
Set up Volume function V(h)
Volume of the cone:
$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2Rh - h^2)h = \frac{1}{3}\pi (2Rh^2 - h^3)$$
Differentiate with respect to h to find critical height
$$\frac{dV}{dh} = \frac{1}{3}\pi (4Rh - 3h^2)$$
Setting $\frac{dV}{dh} = 0 \implies h(4R - 3h) = 0$.
Since height $h \ne 0$, $h = \frac{4R}{3}$.
Second derivative test for maximum
$$\frac{d^2V}{dh^2} = \frac{1}{3}\pi(4R - 6h)$$
At $h = \frac{4R}{3}$:
$$\frac{d^2V}{dh^2} = \frac{1}{3}\pi\left(4R - 6\cdot\frac{4R}{3}\right) = \frac{1}{3}\pi(4R - 8R) = -\frac{4\pi R}{3} < 0$$
Hence, volume is maximum when $h = \frac{4R}{3}$.
Calculate Maximum Volume
$$V_{\max} = \frac{1}{3}\pi \left[ 2R\left(\frac{4R}{3}\right)^2 - \left(\frac{4R}{3}\right)^3 \right] = \frac{1}{3}\pi \left[ \frac{32R^3}{9} - \frac{64R^3}{27} \right] = \frac{1}{3}\pi \left( \frac{32R^3}{27} \right)$$
$$= \frac{8}{27} \left( \frac{4}{3}\pi R^3 \right) = \frac{8}{27} V_{\text{sphere}}$$