如图所示，一个海上油井位于海中的 $W$ 点，距离陆地上的 $A$ 点有 8 km。石油需要输送到距离 $A$ 点 12 km 的 $B$ 点。铺设管道时，先从 $W$ 铺设一条直的水下管道到 $A$ 和 $B$ 之间的某个点 $P$，然后通过另一条铺设在陆地上的直管道输送到 $B$ 点。 如果铺设管道的成本为水下每公里 60,000 美元，陆地上每公里 45,000 美元，那么点 $P$ 距离 $A$ 点应该多远，才能使铺设管道的成本最小？

1. Formulate an equation for the total cost of laying the pipe based on the position of point $P$.

Let the distance $AP$ be $x$ km, where $0 \le x \le 12$.
By Pythagoras' theorem, the length of the underwater pipe is $WP = \sqrt{8^2 + x^2} = \sqrt{x^2 + 64}$ km.
The length of the pipe on land is $PB = 12 - x$ km.
The total cost $C$ (in $) is given by:
$C = 60000\sqrt{x^2 + 64} + 45000(12 - x)$

2. Differentiate the cost function with respect to $x$ to find the stationary point.

To find the minimum cost, we differentiate $C$ with respect to $x$ and set $\frac{dC}{dx} = 0$:
$\frac{dC}{dx} = 60000 \left( \frac{1}{2}(x^2 + 64)^{-\frac{1}{2}} (2x) \right) - 45000$
$\frac{dC}{dx} = \frac{60000x}{\sqrt{x^2 + 64}} - 45000 = 0$

3. Solve the algebraic equation to find the value of $x$.

$\frac{60000x}{\sqrt{x^2 + 64}} = 45000$
Dividing by 15000:
$\frac{4x}{\sqrt{x^2 + 64}} = 3$
$4x = 3\sqrt{x^2 + 64}$
Squaring both sides:
$16x^2 = 9(x^2 + 64)$
$16x^2 = 9x^2 + 576$
$7x^2 = 576$
Since distance $x > 0$,
$x = \sqrt{\frac{576}{7}} = \frac{24}{\sqrt{7}} \approx 9.0714 \text{ km}$

4. Verify that this stationary point results in a minimum cost.

Use the second derivative test:
$\frac{d^2C}{dx^2} = 60000 \left[ \frac{\sqrt{x^2+64} \cdot (1) - x \cdot \left(\frac{x}{\sqrt{x^2+64}}\right)}{x^2+64} \right]$
$\frac{d^2C}{dx^2} = 60000 \left[ \frac{x^2+64 - x^2}{(x^2+64)^{\frac{3}{2}}} \right] = \frac{3840000}{(x^2+64)^{\frac{3}{2}}}$
Since $\frac{d^2C}{dx^2} > 0$ for all valid $x > 0$, the cost $C$ is definitively minimised when $x = \frac{24}{\sqrt{7}} \approx 9.07$ km.