Minimise time of travel (optimisation question)

Question

Okay, need some guidance helping child with math homework. An accident has happened and the ambulance needs to get there in the quickest time. Here is the situation (I added A through D markers):

AMBULANCE              40km
A========================B====================C
                         <-------- x -------->|
                                              |
                                              |
                                              | 15km
                  BUSH                        |
                                              |
                                              |
                                              D ACCIDENT

The ambulance can do 20km/h on the road AC (which is 40km long), and 10km/h in the bush (which is everything except the road).

Since we need to minimise the time taken, there's obviously some point B along the road where it's faster to go off-road straight to D rather than going through C. Let's use x as the BC distance.

So, the time taken to traverse AB and BD (using Pythagoras for the latter) is:

$$\frac{40 - x}{20} + \frac{\sqrt{x^2 + 15^2}}{10}$$

This can be converted into:

$$\frac{40}{20} - \frac{x}{20} + \frac{2\sqrt{x^2 + 15^2}}{20}$$

and further simplified to (since minimising doesn't care about constant additions or common divisors):

$$ 2\sqrt{x^2 + 15^2} - x$$

But I'm not sure where to go from there. Being a computer nerd, I wrote a Python program to calculate the value of x that would deliver the minimum (checking zero through forty in 0.01 increments).

I also think (though from many years ago) that integral calculus may get me the answer as well.

Unfortunately, my child has done neither of those things in class so I'm not sure they're the correct way to go.

What would be the next step, and are there any other (i.e., simpler) ways to work out this problem?

Answer 1

For the Physics-inclined:

$\def\km{\,\mathrm{km}}$ $\def\kmh{\km/\mathrm{h}}$ Light knows the answer and it follows Snell's law:

$$ \frac{\sin\alpha}{\sin\beta}=\frac{20\kmh}{10\kmh}=2 $$ with the angles of incidence

• $\alpha=90^°$ is the angle of direction of ambulance on the road (measured against a road's normal)

• $\beta$ is the angle of direction of ambulance in the bush (measured against a road's normal)

We get $\sin\beta=0.5$ and hence $\beta=\pi/6=30^°$. This means angle $\angle CBD = 60°$ and therefore $$ \frac{15\km}x = \tan 60° = \sqrt3 \quad\Rightarrow \quad x=\frac1{\sqrt3}\cdot 15\km = 5\sqrt3\km \approx 8.66\km. $$

Because $x\leqslant 40\km$, this is the result. (If $x\geqslant40\km$ then the best path would be not to use the road at all, and to go 100% bushes.)

Answer 2

Making the problem more general, you want to minimize $$f(x)=2\sqrt{x^2 + a^2} - x$$ for which $$f'(x)=\frac{2 x}{\sqrt{x^2+a^2}}-1\qquad \text{and} \qquad f''(x)=\frac{2 a^2}{\left(a^2+x^2\right)^{3/2}}$$ So, the first derivative cancels when $$2x=\sqrt{x^2+a^2}\implies 4x^2=x^2+a^2\implies ???$$

Answer 3

Differentiate your expression for time: $T = \frac{40-x}{20}+\frac{\sqrt{x^2+15^2}}{10}$

Leading to $T' = -\frac{1}{20} + \frac{x}{10\sqrt{x^2+15^2}}$

Extreme values can be found by setting $T'=0~~~~==>~~2x = \sqrt{x^2+15^2}$

Solve for x.