[Cambridge/N75/II/5] A cylinder of radius $x$ is inscribed in a fixed sphere of internal radius $R$ such that the circumferences of the circular ends of the cylinder are in contact with the inner surface of the sphere. Show that $A^2 = 16 ext{\pi}^2 x^2 (R^2 - x^2)$, where $A$ is the curved surface area of the cylinder. Prove that, if $x$ is made to vary, the maximum value of $A$ is obtained when the height of the cylinder is equal to its diameter.

1. Relate the height of the cylinder to its radius $x$ and the sphere's radius $R$ using Pythagoras' Theorem.

Let the height of the inscribed cylinder be $h$.
Considering the cross-section passing through the center of the sphere, the distance from the sphere's center to the center of the cylinder's circular end is $\frac{h}{2}$.
By Pythagoras' Theorem on the right-angled triangle formed:
$x^2 + \left(\frac{h}{2}\right)^2 = R^2$
$\frac{h^2}{4} = R^2 - x^2$
$h^2 = 4(R^2 - x^2)$

2. Express the square of the curved surface area, $A^2$, in terms of $x$ and $R$.

The curved surface area of the cylinder is $A = 2\pi x h$.
Squaring both sides gives:
$A^2 = 4\pi^2 x^2 h^2$
Substituting $h^2 = 4(R^2 - x^2)$:
$A^2 = 4\pi^2 x^2 [4(R^2 - x^2)]$
$A^2 = 16\pi^2 x^2(R^2 - x^2)$ (Shown)

3. Differentiate $A^2$ with respect to $x$ to find the stationary point for maximum area.

Let $Z = A^2 = 16\pi^2 R^2 x^2 - 16\pi^2 x^4$.
Differentiating $Z$ with respect to $x$:
$\frac{dZ}{dx} = 32\pi^2 R^2 x - 64\pi^2 x^3$
To find the maximum, set $\frac{dZ}{dx} = 0$:
$32\pi^2 x (R^2 - 2x^2) = 0$
Since the radius $x > 0$, we divide by $32\pi^2 x$:
$R^2 - 2x^2 = 0 \implies x^2 = \frac{R^2}{2}$

4. Perform the second derivative test to confirm the stationary point is a maximum.

Differentiate $\frac{dZ}{dx}$ again with respect to $x$:
$\frac{d^2Z}{dx^2} = 32\pi^2 R^2 - 192\pi^2 x^2$
Substitute $x^2 = \frac{R^2}{2}$ into the second derivative:
$\frac{d^2Z}{dx^2} = 32\pi^2 R^2 - 192\pi^2 \left(\frac{R^2}{2}\right)$
$\frac{d^2Z}{dx^2} = 32\pi^2 R^2 - 96\pi^2 R^2 = -64\pi^2 R^2$
Since $R > 0$, $\frac{d^2Z}{dx^2} < 0$.
Therefore, $Z$ (and hence $A$, since $A>0$) is maximum when $x^2 = \frac{R^2}{2}$.

5. Compare the height and the diameter of the cylinder at this maximum.

At the maximum $A$, we have $x^2 = \frac{R^2}{2}$.
The diameter of the cylinder is $D = 2x$, so:
$D^2 = 4x^2 = 4\left(\frac{R^2}{2}\right) = 2R^2$
The square of the height is:
$h^2 = 4(R^2 - x^2) = 4\left(R^2 - \frac{R^2}{2}\right) = 2R^2$
Comparing the two, we get $D^2 = h^2$.
Since lengths $D$ and $h$ must be positive, it follows that $h = D$.
Thus, the maximum value of $A$ is obtained when the height of the cylinder is equal to its diameter. (Proved)