[已知半径为 $r$ 的球体的表面积和体积分别为 $4\pi r^2$ 和 $\frac{4}{3}\pi r^3$。] 一家制造商生产障碍物，该障碍物由一个半径为 $r\text{ cm}$ 的半球与一个底面半径为 $r\text{ cm}$、高为 $h\text{ cm}$ 的圆柱体拼接而成（见图）。 制造商决定每个障碍物的外表面积（不包括底面）为 $16200\text{ cm}^2$。
利用微分，求障碍物最大体积的精确值。

1. Formulate an equation for the external surface area excluding the base.

The external surface area consists of the curved surface area of the cylinder and the curved surface area of the hemisphere.
$2\pi rh + 2\pi r^2 = 16200$

2. Express the height $h$ in terms of $r$.

$2\pi rh = 16200 - 2\pi r^2$
$h = \frac{16200 - 2\pi r^2}{2\pi r}$
$h = \frac{8100}{\pi r} - r$

3. Formulate an equation for the volume of the barrier in terms of $r$ and $h$.

The volume $V$ is the sum of the volumes of the cylinder and the hemisphere:
$V = \pi r^2 h + \frac{2}{3}\pi r^3$

4. Express the volume $V$ entirely in terms of $r$.

$V = \pi r^2 \left( \frac{8100}{\pi r} - r \right) + \frac{2}{3}\pi r^3$
$V = 8100r - \pi r^3 + \frac{2}{3}\pi r^3$
$V = 8100r - \frac{1}{3}\pi r^3$

5. Differentiate $V$ with respect to $r$.

$\frac{\mathrm{d}V}{\mathrm{d}r} = 8100 - \pi r^2$

6. Equate the derivative to zero to find the stationary value of $r$.

For maximum volume, $\frac{\mathrm{d}V}{\mathrm{d}r} = 0$:
$8100 - \pi r^2 = 0$
$\pi r^2 = 8100$
$r^2 = \frac{8100}{\pi}$
$r = \frac{90}{\sqrt{\pi}} \quad \text{(since } r > 0 \text{)}$

7. Verify that this value of $r$ gives a maximum volume using the second derivative test.

$\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -2\pi r$
When $r = \frac{90}{\sqrt{\pi}}$, $\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -2\pi \left( \frac{90}{\sqrt{\pi}} \right) < 0$.
Since the second derivative is negative, the volume is maximized.

8. Calculate the exact maximum volume.

$V = 8100 \left( \frac{90}{\sqrt{\pi}} \right) - \frac{1}{3}\pi \left( \frac{90}{\sqrt{\pi}} \right)^3$
$V = \frac{729000}{\sqrt{\pi}} - \frac{1}{3}\pi \left( \frac{729000}{\pi\sqrt{\pi}} \right)$
$V = \frac{729000}{\sqrt{\pi}} - \frac{243000}{\sqrt{\pi}}$
$V = \frac{486000}{\sqrt{\pi}}$