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: Alternative: maximizing V^2 to avoid the square root:
Since \(V\) and \(V^2\) are maximized at the same \(h\) (V>0), consider \(V^2=\pi^2h^2\big(R^2-h^2/4\big)^2\), or more simply keep working with \(r^2\) as the free variable instead of \(h\).

Step 2: Expressing V in terms of r^2 = t:
With \(h=2\sqrt{R^2-r^2}\), \(V=2\pi r^2\sqrt{R^2-r^2}\). Let \(t=r^2\): \(V^2=4\pi^2t^2(R^2-t)\).

Step 3: Maximizing the polynomial V^2 in t:
\(\dfrac{d}{dt}\big[t^2(R^2-t)\big]=2tR^2-3t^2=t(2R^2-3t)=0\Rightarrow t=\dfrac{2R^2}{3}\) (rejecting \(t=0\)).

Step 4: Recovering h from this t:
\(r^2=t=\dfrac{2R^2}{3}\Rightarrow h^2=4(R^2-t)=4\Big(R^2-\dfrac{2R^2}{3}\Big)=\dfrac{4R^2}{3}\Rightarrow h=\dfrac{2R}{\sqrt3}\), the same result.

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