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: Parametrize by the base radius r instead of the height, as an alternate route:
From $r^2+h^2=R^2$, write $h=\sqrt{R^2-r^2}$, so $H=2h=2\sqrt{R^2-r^2}$ and $V=\pi r^2\cdot2\sqrt{R^2-r^2}$.

Step 2: Maximize $V^2$ instead (avoids the square root in differentiation) — same maximizer since V, r ≥ 0:
$V^2=4\pi^2r^4(R^2-r^2)=4\pi^2(R^2r^4-r^6)$. $\dfrac{d(V^2)}{dr}=4\pi^2(4R^2r^3-6r^5)=4\pi^2r^3(4R^2-6r^2)=0$.

Step 3: Solve the resulting equation for r (excluding the trivial r=0):
$4R^2-6r^2=0\Rightarrow r^2=\dfrac{2R^2}{3}$.

Step 4: Back-substitute to find h and H:
$h^2=R^2-r^2=R^2-\dfrac{2R^2}{3}=\dfrac{R^2}{3}\Rightarrow h=\dfrac{R}{\sqrt3}$, so $H=2h=\dfrac{2R}{\sqrt3}$, matching the direct approach. Volume: $V=\pi\cdot\dfrac{2R^2}{3}\cdot\dfrac{2R}{\sqrt3}=\dfrac{4\pi R^3}{3\sqrt3}$.

Final Answer:
\[ \boxed{H=2R/\sqrt3,\quad V_{max}=4\pi R^3/(3\sqrt3)} \]