图示一个湖泊（阴影部分），其边缘有一条笔直平坦的道路 AXB。点 C 在湖面上，B 是距离 C 最近的点。距离 AB 为 30 m，BC 为 20 m。 纳塔莉（Natalie）在道路上的 A 点，想要尽快到达 C 点。她的奔跑速度为 4 m/s，游泳速度为 2 m/s。 她意识到，为了使到达 C 的总时间最短，必须在 A 和 B 之间的某个点 X 离开道路，然后直接游向 C。 设 AX 为 $x$ m，$T$ 为纳塔莉从 A 到 X 再从 X 到 C 所用的总时间（以秒为单位）。

1. Find the distance $XC$ using Pythagoras' theorem

Since $B$ is the point on the road closest to $C$, $\angle XBC = 90^\circ$.
$XB = AB - AX = 30 - x$
Using Pythagoras' theorem in $\triangle XBC$:
$XC^2 = XB^2 + BC^2$
$XC^2 = (30 - x)^2 + 20^2$
$XC^2 = 900 - 60x + x^2 + 400$
$XC = \sqrt{x^2 - 60x + 1300}$

2. Express the total time $T$ in terms of $x$

Time taken to run from $A$ to $X$ = $\frac{\text{Distance } AX}{\text{Speed}} = \frac{x}{4}$
Time taken to swim from $X$ to $C$ = $\frac{\text{Distance } XC}{\text{Speed}} = \frac{\sqrt{x^2 - 60x + 1300}}{2}$
Total time $T = \frac{x}{4} + \frac{1}{2}\sqrt{x^2 - 60x + 1300}$ (Shown)

1. Differentiate the expression for $T$ with respect to $x$ using the chain rule

$T = \frac{1}{4}x + \frac{1}{2}(x^2 - 60x + 1300)^{\frac{1}{2}}$

$\frac{\mathrm{d}T}{\mathrm{d}x} = \frac{1}{4} + \frac{1}{2} \left[ \frac{1}{2}(x^2 - 60x + 1300)^{-\frac{1}{2}} \cdot (2x - 60) \right]$

2. Simplify the derivative

$\frac{\mathrm{d}T}{\mathrm{d}x} = \frac{1}{4} + \frac{1}{4}(2x - 60)(x^2 - 60x + 1300)^{-\frac{1}{2}}$

$\frac{\mathrm{d}T}{\mathrm{d}x} = \frac{1}{4} + \frac{x - 30}{2\sqrt{x^2 - 60x + 1300}}$

1. Find the derivative of $T$ with respect to $x$ (from part b)

$\frac{\mathrm{d}T}{\mathrm{d}x} = \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{2\sqrt{x^2 - 60x + 1300}} \cdot (2x - 60)$
$\frac{\mathrm{d}T}{\mathrm{d}x} = \frac{1}{4} + \frac{x - 30}{2\sqrt{x^2 - 60x + 1300}}$

2. Set the derivative to zero to find the minimum time

$\frac{1}{4} + \frac{x - 30}{2\sqrt{x^2 - 60x + 1300}} = 0$
$\frac{30 - x}{2\sqrt{x^2 - 60x + 1300}} = \frac{1}{4}$

3. Rearrange and square both sides to form a quadratic equation

$2(30 - x) = \sqrt{x^2 - 60x + 1300}$
$4(30 - x)^2 = x^2 - 60x + 1300$
$4(900 - 60x + x^2) = x^2 - 60x + 1300$
$3600 - 240x + 4x^2 = x^2 - 60x + 1300$
$3x^2 - 180x + 2300 = 0$

4. Solve the quadratic equation for $x$ and select the valid root

$x = \frac{180 \pm \sqrt{(-180)^2 - 4(3)(2300)}}{2(3)}$
$x = \frac{180 \pm \sqrt{32400 - 27600}}{6}$
$x = \frac{180 \pm \sqrt{4800}}{6}$
$x = 30 \pm \frac{20\sqrt{3}}{3}$
Since $X$ is between $A$ and $B$, $x < 30$, so we reject $x = 30 + \frac{20\sqrt{3}}{3}$.
$x = 30 - \frac{20\sqrt{3}}{3} \approx 18.453$
$x \approx 18.5$ (to 3 s.f.)