# Optimization problem with a round lake

by frosty8688
Tags: lake, optimization
 P: 126 The rowing distance is about 3.46.
 P: 126 The arc length is 2.09.
 P: 126 It would take her over two hours to row and walk. So she should row, which would take 2 hours.
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Thanks
P: 9,095
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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P: 4,128
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
 P: 126 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.
 P: 126 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.
 P: 126 I agree that walking is the fastest route.
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