The diagram shows a lake (shaded) bordered by a straight level road AXB. Point C is on the surface of the lake and B is the point closest to C. The distance AB is 30 m and BC is 20 m. Natalie is at point A on the road and wants to get to point C as quickly as possible. She can run at a speed of 4 m/s and swim at a speed of 2 m/s. She realizes that in order to minimize the total time to get to C, it is necessary to leave the road at some point X between A and B, and then swim directly to C. Let AX be $x$ m and $T$ be the total time, in seconds, taken by Natalie to get from A to X and then from X to C.

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.)