The diagram shows a solid object. It is made up of a cylinder with a hemisphere above it. The cylinder has a radius of $x\text{ cm}$ and a height of $y\text{ cm}$. The radius of the hemisphere is $2x\text{ cm}$.

1. Formulate an equation for the total volume of the solid.

Volume of cylinder $= \pi x^2 y$
Volume of hemisphere $= \frac{2}{3}\pi(2x)^3 = \frac{16}{3}\pi x^3$
Total volume $= \pi x^2 y + \frac{16}{3}\pi x^3 = 72\pi$

2. Make $y$ the subject of the formula.

Divide the entire equation by $\pi$:
$x^2 y + \frac{16}{3}x^3 = 72$
$x^2 y = 72 - \frac{16}{3}x^3$
$y = \frac{72}{x^2} - \frac{16}{3}x$

1. Find the expression for $y$ in terms of $x$ from the volume constraint (from part a)

Volume of cylinder $= \pi x^2 y$
Volume of hemisphere $= \frac{2}{3}\pi (2x)^3 = \frac{16}{3}\pi x^3$
Total volume $= \pi x^2 y + \frac{16}{3}\pi x^3 = 72\pi$
$x^2 y + \frac{16}{3}x^3 = 72$
$y = \frac{72}{x^2} - \frac{16}{3}x$

2. Formulate the expression for the total surface area of the solid

Curved surface area of hemisphere $= 2\pi(2x)^2 = 8\pi x^2$
Exposed flat base of hemisphere $= \pi(2x)^2 - \pi x^2 = 3\pi x^2$
Curved surface area of cylinder $= 2\pi x y$
Base of cylinder $= \pi x^2$
Total surface area $A = 8\pi x^2 + 3\pi x^2 + 2\pi x y + \pi x^2 = 12\pi x^2 + 2\pi x y$

3. Substitute $y$ into the surface area expression and simplify to show the required form

$A = 12\pi x^2 + 2\pi x \left( \frac{72}{x^2} - \frac{16}{3}x \right)$
$A = 12\pi x^2 + \frac{144\pi}{x} - \frac{32}{3}\pi x^2$
$A = \frac{144\pi}{x} + \frac{4}{3}\pi x^2$
$A = 144\pi \left( \frac{1}{x} + \frac{4}{3 \times 144}x^2 \right)$
$A = 144\pi \left( \frac{1}{x} + \frac{x^2}{108} \right)$ (shown)

1. Differentiate $A$ with respect to $x$

$A = 144\pi\left(x^{-1} + \frac{x^2}{108}\right)$
$\frac{\mathrm{d}A}{\mathrm{d}x} = 144\pi\left(-x^{-2} + \frac{2x}{108}\right)$
$\frac{\mathrm{d}A}{\mathrm{d}x} = 144\pi\left(-\frac{1}{x^2} + \frac{x}{54}\right)$

2. Set the derivative to zero to find the stationary value of $x$

For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}x} = 0$
$144\pi\left(-\frac{1}{x^2} + \frac{x}{54}\right) = 0$
$\frac{x}{54} = \frac{1}{x^2}$
$x^3 = 54$
$x = \sqrt[3]{54} \approx 3.77976$

3. Determine the nature of the stationary value using the second derivative test

$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 144\pi\left(2x^{-3} + \frac{1}{54}\right)$
$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 144\pi\left(\frac{2}{x^3} + \frac{1}{54}\right)$
When $x = \sqrt[3]{54}$, $x > 0$, so $\frac{\mathrm{d}^2A}{\mathrm{d}x^2} > 0$.
Since the second derivative is positive, the stationary value is a minimum.