## Optimization problem with a round lake

 Quote by Simon Bridge Hmm ... that means that she has travelled a total of ##2\frac{2}{3}+1\frac{1}{3}=4\text{miles}## ... which is the diameter of the circle. But that would only happen by rowing the whole way (since rowing is the only direct way across the lake)! Does this make sense? Can she walk on water? I cannot tell where you made the mistake because you have not answered my question: what distance does x represent? For example - if we make the center of the lake the origin O of a Cartesian coordinate system, so that A=(0,R), C=(0,-R), and B=(x,y) is guessed to be in the first quadrant ... then the distance rowed is ##d=|AB| = \sqrt{x^2+(y-R)^2}## and the equation of the lake-shore is ##x^2+y^2=R^2##. If ##\angle AOB = \theta## then the distance walked is given by ##s=(\pi-\theta)R## I'm not sure how you can get the arclength s easily in terms of rectangular coordinates, but you can get the rowing distance d in terms of ##\theta## using the cosine rule. None of these equations look anything like yours ... so x does not appear to be a rectangular coordinate. So what is it?
I see your point. I was just trying to do it a different way.
 The rowing distance is about 3.46.
 The arc length is 2.09.
 It would take her over two hours to row and walk. So she should row, which would take 2 hours.

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Which one: I made a number of them. But I feel that if you really saw the point you would have answered the questions!
 I was just trying to do it a different way.
Yes: clearly - and you still neglect to tell me which way that was!
 The rowing distance is about 3.46. The arc length is 2.09.
How are you getting these values?
 It would take her over two hours to row and walk. So she should row, which would take 2 hours.
Nonsense!

If you had set up your equation correctly, and rowing the whole way was the quickest route, then there would be a minima in your equation for that situation.
There wasn't - instead the minima for the equation gave a slower route that the slowest possible path (see below) - therefore, your equation is set up incorrectly.

If she rowed the whole way, then, yes: she would take ##R/u=4/2## or 2hr to get there. This is the slowest route of the ones available. For instance: if she walked the whole way, then she would take ##\pi R/v = 2\pi/4 = \pi/2## or 1.57 hr to get there ... so, of the two choices she should walk - surely?

However, there may still be some advantage in rowing some of the way.

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When I do it I find there is an even slower route (corresponding to a maximum time): row to point B, where the angle BOC is ___ radians, then walk the rest of the way, for a total time of 2.2556 hours; I am keeping the answer blank in order to leave something for the OP to do. Thus, rowing the whole way is not the slowest route. When done properly (in terms of the angle θ = BOC) the time function T(θ) has local minima at θ = 0 (row all the way) and θ = π (walk all the way) and a global maximum at an intermediate point.

RGV
 I got the same answer as regards to rowing and walking times. I figured out that the chord AB is approximately 1 unit or 1 mile away from the point O and the radius is 2 miles. So the chord is 2$\sqrt{r^{2}-d^{2}}$ with d being the distance of the chord from point O. The answer for the chord length is 2$\sqrt{3}$ which equals about 3.46. I divided this by 2 and it came out to be 1.73 hours of rowing.
 I then took 2 arcsin of (cL/2Ra). cL is chord length and Ra is radius and ended up with $\frac{2}{3}$∏. This is the arc length. I then divided this by 4 and it came out to be about 0.52.
 I agree that walking is the fastest route.

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 I figured out that the chord AB is approximately 1 unit or 1 mile away from the point O
How? Presumably that is for the maxima?

You are right though - making x the distance from O to the center of the chord (if that is what you did) is an interesting way of doing it.
(I messed up my description before - thanks Ray.)