Question

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

Hint:

Express V as a function of h alone using the sphere constraint, then maximize with calculus.

Answer 1

Step 1: Setting up the volume in terms of one variable:
Let the cylinder have height \(h\) and radius \(r\). Since it is inscribed in the sphere, \(r^{2}+\Big(\dfrac{h}{2}\Big)^{2}=R^{2}\ \Rightarrow\ r^{2}=R^{2}-\dfrac{h^{2}}{4}\).

Step 2: Writing the volume as a function of h alone:
\(V=\pi r^{2}h=\pi h\Big(R^{2}-\dfrac{h^{2}}{4}\Big)=\pi R^{2}h-\dfrac{\pi h^{3}}{4}\).

Step 3: Differentiating and setting to zero:
\(\dfrac{dV}{dh}=\pi R^{2}-\dfrac{3\pi h^{2}}{4}=0\ \Rightarrow\ h^{2}=\dfrac{4R^{2}}{3}\ \Rightarrow\ h=\dfrac{2R}{\sqrt3}\).

Step 4: Confirming it is a maximum:
\(\dfrac{d^{2}V}{dh^{2}}=-\dfrac{3\pi h}{2}<0\) for \(h>0\), confirming this critical point gives the maximum volume.

Final Answer:
\[ \boxed{h=\dfrac{2R}{\sqrt3}} \]