Question

Prove that the height of the cylinder of maximum volume inscribed in a sphere of radius \(R\) is \(\dfrac{2R}{\sqrt3}\). Also find the maximum volume of the cylinder.

Hint:

Express V as a function of the half-height h using r^2+h^2=R^2, then maximize via dV/dh=0.

Answer 1

Step 1: Set up the constraint between cylinder half-height and radius:
Let the cylinder have height \(H=2h\) and base radius \(r\). Since the cylinder is inscribed in a sphere of radius \(R\), the center of the sphere, the center of the cylinder's axis and a top-rim point form a right triangle: \(r^2+h^2=R^2\).

Step 2: Write the volume as a function of \(h\) alone:
\[ V=\pi r^2H=\pi(R^2-h^2)(2h)=2\pi(R^2h-h^3) \]

Step 3: Differentiate and set to zero:
\[ \dfrac{dV}{dh}=2\pi(R^2-3h^2)=0\implies h^2=\dfrac{R^2}3\implies h=\dfrac R{\sqrt3} \]
Check it's a maximum: \(\dfrac{d^2V}{dh^2}=2\pi(-6h)<0\) for \(h>0\), confirming a maximum.

Step 4: Find the height and the maximum volume:
\[ H=2h=\dfrac{2R}{\sqrt3}\quad\text{(as required)} \]
\[ r^2=R^2-h^2=R^2-\dfrac{R^2}3=\dfrac{2R^2}3 \]
\[ V_{max}=\pi r^2H=\pi\cdot\dfrac{2R^2}3\cdot\dfrac{2R}{\sqrt3}=\dfrac{4\pi R^3}{3\sqrt3}=\dfrac{4\sqrt3\,\pi R^3}9 \]

Final Answer:
\[ \boxed{H=\dfrac{2R}{\sqrt3},\qquad V_{max}=\dfrac{4\pi R^3}{3\sqrt3}} \]