# Difficult Max/Min problem, calculating least distance.

"A man can row at 3 m.p.h. and run at 5 m.p.h. He is 5 miles out to sea and wishes to get to a point on the coast 13 miles from where he is now.Where should he land on the coast to get there as soon as possible?"

I've been on this for hours now and I'm assuming it's something like this: ac = 5 miles

ab = 13 miles

Am I on the right track? am I to find the minimum distance of bc? I tried using the Velocity = distance/time equation and plugging it into the pythagoras theorem but still couldn't figure it out..if anyone has some hints I'd be grateful.

Note here that you are asked to minimize the time of the trip. You would decompose the problem into something like $$T(p_1 \cup p_2)=T(p_1)+T(p_2)$$ where $p_1$ would be the sea path, and $p_2$ the land path. You would need to minimize their total time. Now, you take a point $P$ on the coast (i.e. BC). Supposing point a was the origin, we place $P=(x,y)$

Let's call the distance from C to P d. d is also $P$'s y coordinate. Then $BP=BC-d$. Now, your time of sea would be $$T(p_1)=\frac{\sqrt{5^2+d^2}}{3}$$ using the formula $$V=\frac{d}{t}$$ Now symmetrically, $$T(p_2)=\frac{(12-d)}{5}$$ Since $BC=\sqrt{13^2-5^2}$ Now you end up with $$T(p_1 \cup p_2)=\frac{\sqrt{5^2+d^2}}{3}+\frac{(12-d)}{5}$$. $T(p_1 \cup p_2)$ is the function you want to minimize. Consequently, you solve $$\frac{\partial(T(p_1 \cup p_2))}{\partial d} =0$$ After this, you should get the max/min of the equation, which you will first test as being the right one, and only then use in your answer.

• I'm so annoyed I didn't crack this by myself...thanks very much though. Jun 18, 2015 at 21:30

What you need to do is to let $(x,y)$ be the point at which he hits the shore, stopping rowing and starting running (assuming he just hops out of his boat and takes off).

This will be controlled by the angle $\theta$ he makes. Clearly, if he goes along segment $ac$, it will take him 4 hours and 20 minutes--13miles / 3mph = 4 hrs, 20 min. This angle would be $\arccos\left(\frac{5}{12}\right)$.

If he goes along segment $ac$, this will take him 5 / 3 = 1 hr 40 min + 12 / 5 = 2 hr, 24 min, which is 4hrs, 4 minutes. Let's just call this angle 0.

The time equation is then given by $t(\theta) = 5\sec(\theta) / 3 + (12-5\cot(\theta))/ 5$ (this may not be correct; I don't currently have access to any scratch paper to really nail this down)

Then you simply need to apply your max/min problem-solving skills. This is equivalent to Ghost's above, but his solution is somewhat simpler, and doesn't involve messy trig calculations, so I would use his--with the slight modification that you need to divide by the speed instead of multiplying: $T(p_1 \cup p_2) = \sqrt{5^2 + d^2} / 3 + (12 -d ) /5$ However, this works just as well. Interestingly, light faces the same problem when it has to cross different medium, and light solves the problem by following snell's lay... You could too.

Since you ask for a hint, this is all I can say. Just realize that the person would try to land higher on the coastline till the gain in the form of shortcut is cancelled by the otherwise longer route at higher speed on land. This would lead you to proving snell's ratio which highly depends upon the speeds of person in the two medium...