Question

Prove that cone of the maximum volume inscribed in a sphere of radius $R$ is $\tfrac{8}{27}$ of the volume of the sphere.

Hint:

The maximum volume cone inscribed in a sphere always has height $\tfrac{4R}{3}$. Its volume turns out to be $\tfrac{8}{27}$ of the sphere's volume.

Answer 1

Step 1: Consider the geometry. Let a right circular cone of height $h$ and base radius $r$ be inscribed in a sphere of radius $R$. - Vertex of cone at top of sphere. - Base of cone inside sphere. By Pythagoras (in right-angled triangle from sphere center to cone base): \[ r^2 + \left(R-h\right)^2 = R^2 \] \[ r^2 = 2Rh - h^2 \]

Step 2: Write volume of cone. \[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2Rh - h^2)h \] \[ V = \frac{\pi}{3}(2Rh^2 - h^3) \]

Step 3: Maximize volume. Differentiate: \[ \frac{dV}{dh} = \frac{\pi}{3}(4Rh - 3h^2) \] Set $\frac{dV}{dh}=0$: \[ 4Rh - 3h^2 = 0 \implies h(4R - 3h)=0 \] So, $h = 0$ or $h=\tfrac{4R}{3}$. We take $h=\tfrac{4R}{3}$.

Step 4: Compute $r$. \[ r^2 = 2R\left(\tfrac{4R}{3}\right) - \left(\tfrac{4R}{3}\right)^2 \] \[ = \tfrac{8R^2}{3} - \tfrac{16R^2}{9} = \tfrac{8R^2}{9} \] So, $r = \tfrac{2\sqrt{2}R}{3}$.

Step 5: Maximum volume of cone. \[ V_{\max} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \cdot \frac{8R^2}{9} \cdot \frac{4R}{3} \] \[ = \frac{32}{81}\pi R^3 \]

Step 6: Volume of sphere. \[ V_{\text{sphere}} = \frac{4}{3}\pi R^3 \]

Step 7: Ratio. \[ \frac{V_{\max}}{V_{\text{sphere}}} = \frac{\tfrac{32}{81}\pi R^3}{\tfrac{4}{3}\pi R^3} = \frac{32}{81}\cdot\frac{3}{4} = \frac{8}{27} \]

Final Answer: \[ \boxed{\;\;V_{\max} = \frac{8}{27}\,V_{\text{sphere}}\;\;} \]