Math Is Fun Forum

headhurts

Member

Registered: 2014-05-04

Posts: 8

Optimization Problem

Good old rowing vs walking optimization problem.

In this case, we have a circular lake with radius 2. Walk 6kmph, row 2kmph. Go to the opposite side of the lake.

Say I walk from point A along the circumference to point B, then row across to destination C.

Since the three point on the semicircle will make a right angle, I can compute BC in terms of 'theta' (angle CAB).

But how do I compute the distance AC (perimeter of the wedge) in terms of 'theta'? In other words, how do I compute the angle AOB in terms of 'theta', since I need the angle AOB to compute the perimeter of the wedge.

headhurts

Member

Registered: 2014-05-04

Posts: 8

Re: Optimization Problem

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Optimization Problem

hi headhurts,

I'm wondering if there is some information missing here.  I've assumed the task is to get from A to C, by walking first to B and then rowing to C, as quickly as possible. 

AOB is isosceles so S = pi - 2T  (I'm using radians)

Arc length AB = Sr = 2(pi - 2T)

BC = 4sin(T)

So total time = 2(pi-2T)/6 + 4sin(T)/2 = (pi-2T)/3 + 2sin(T)

I've graphed this here:

http://www.mathsisfun.com/data/function … /3+2sin(x)

Once you've focussed in on acceptable values it looks like the optimum is T = 0, ie walk it all.  ???

Please consider this and post back,

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

headhurts

Member

Registered: 2014-05-04

Posts: 8

Re: Optimization Problem

AOB is isosceles so S = pi - 2T  (I'm using radians)

Thank you!!!

Couldn't quite figure out that relationship before you mentioned isosceles triangle. So many relationships to remember. D: