一个无盖的金属水箱呈直圆柱体形状。制造该水箱所用金属的总成本为固定金额 $A。用于制作底面的金属每单位面积成本为 $B，用于制作侧面的金属每单位面积成本为 $C，其中 B 和 C 为常数。
已知水箱底面半径为 $r$，用 $r$、$A$、$B$ 和 $C$ 表示水箱的体积 $V$。
利用微分法求，当水箱体积最大时，用于制作底面的金属成本（用 $A$ 表示）。

1. Formulate the equation for total cost $A$ and express height $h$ in terms of $r$, $A$, $B$, and $C$.

Let the height of the cylinder be $h$.
Area of base = $\pi r^2$
Area of curved surface = $2\pi r h$
Total cost, $A = B(\pi r^2) + C(2\pi r h)$
$2\pi r h C = A - B\pi r^2$
$h = \dfrac{A - B\pi r^2}{2\pi r C}$

2. Express the volume $V$ of the tank in terms of $r$, $A$, $B$, and $C$.

Volume of a cylinder, $V = \pi r^2 h$
$V = \pi r^2 \left( \dfrac{A - B\pi r^2}{2\pi r C} \right)$
$V = \dfrac{r(A - B\pi r^2)}{2C}$
$V = \dfrac{Ar - B\pi r^3}{2C}$

3. Differentiate $V$ with respect to $r$ and equate to zero to find the condition for maximum volume.

$\dfrac{\mathrm{d}V}{\mathrm{d}r} = \dfrac{A - 3B\pi r^2}{2C}$
For maximum volume, $\dfrac{\mathrm{d}V}{\mathrm{d}r} = 0$.
$\dfrac{A - 3B\pi r^2}{2C} = 0$
$A - 3B\pi r^2 = 0$

4. Perform the second derivative test to justify that the volume is a maximum.

$\dfrac{\mathrm{d}^2V}{\mathrm{d}r^2} = \dfrac{-6B\pi r}{2C} = -\dfrac{3B\pi r}{C}$
Since the constants $B$ and $C$ represent costs, and $r$ is a length, $B, C, r > 0$.
Therefore, $\dfrac{\mathrm{d}^2V}{\mathrm{d}r^2} < 0$, which confirms that the volume is a maximum.

5. Find the cost of the base in terms of $A$ at the maximum volume.

From $A - 3B\pi r^2 = 0$, we have:
$3B\pi r^2 = A$
$B\pi r^2 = \dfrac{A}{3}$
Since the cost of the base is represented by $B(\pi r^2)$, the cost of the base is $\dfrac{A}{3}$.