一家制造商想设计一个开口圆柱形水箱，如图所示。这个水箱有底但没有顶部。水箱外表面的固定面积为 $600\pi\text{ cm}^2$。水箱的半径 $r$ cm 和高度 $h$ cm 可以变化。

(a) 证明水箱的体积 $V\text{ cm}^3$ 为

(b) 求使 $V$ 取得最大值时对应的 $r$ 的精确值。

(c) 由此，求 $V$ 的最大值。

开口水箱的外表面积由一个底面和一个侧面组成。用表面积条件把 $h$ 表成 $r$，再代入体积公式。

The area of the base is $\pi r^2$, and the curved surface has area $2\pi rh$. Hence

The volume is therefore

对 (a) 的体积式求导。找到驻点后检查导数符号，确认体积先增后减。

Since $r>0$ and $h>0$, the feasible interval is $0<r<\sqrt{600}$. Differentiating gives

The derivative is zero when $r^2=200$, so the only feasible stationary point is $r=10\sqrt2$. The derivative is positive before this point and negative after it throughout the feasible interval. Thus $V$ is maximised at $r=10\sqrt2\text{ cm}$.

也可以直接从表面积条件找最大值。把 $r^2+2rh$ 看成三个正数之和，利用算术平均数与几何平均数的不等式限制体积。

The three positive terms $r^2$, $rh$ and $rh$ have sum $600$. Their product is $r^4h^2=(V/\pi)^2$. By the arithmetic mean–geometric mean inequality,

with equality exactly when $r^2=rh$, that is, when $r=h$. This is feasible and gives $3r^2=600$. Since $r>0$, the maximum occurs at $r=10\sqrt2\text{ cm}$.

把 (b) 求出的最佳半径代回 (a) 的体积式，保留精确值到最后。

At the maximising radius $r=10\sqrt2$ found in (b), $r^2=200$. Therefore