图示一个由薄塑料片制成的容器，形状为中空圆柱体。容器的长度为 $y$ cm，每个半圆形两端的半径均为 $r$ cm。所用塑料的总外表面积为 8500 cm$^2$。

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

$\frac{dV}{dr} = 4250 - \frac{3}{2}\pi r^2$

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

For a stationary value, $\frac{dV}{dr} = 0$:
$4250 - \frac{3}{2}\pi r^2 = 0$
$\frac{3}{2}\pi r^2 = 4250$
$r^2 = \frac{8500}{3\pi}$
$r = \sqrt{\frac{8500}{3\pi}} \approx 30.033$

3. Calculate the stationary value of $V$

$V = 4250(30.033) - \frac{1}{2}\pi (30.033)^3$
$V \approx 85093.9 \approx 85100 \text{ (to 3 s.f.)}$

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

$\frac{d^2V}{dr^2} = -3\pi r$
Since $r > 0$, $\frac{d^2V}{dr^2} < 0$.
Therefore, the stationary value is a maximum.