tricky math problem

The original question and the work I did on it is attached (the work I did is a bit short-cut since I've verified my answer until that point).

My problem is with part B, when I solve for it, I get x = 4/3, which would technically make this traveler travel a bit further than needed to get to the destination, which would make no sense, unless he just really loves to row/walk...

Any guidance on the matter?

And to the mods and experts who will be complaining about providing direct solutions, this is no different than me posting C++ code and having other experts rip it apart and show me better ways of doing the same thing, which does not violate any rules.
ZenotureAsked:
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ozoCommented:
In the real world, walk time would be abs|1-x|/5, not (1-x)/5
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ZenotureAuthor Commented:
Oops, forgot to attach the image. Here it is.
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ZenotureAuthor Commented:
Upload script isn't working for me for some reason... Here are links to the images:

http://zenoture.net/math_prob.png
http://zenoture.net/math_prob_work.png
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sshah254Commented:
No attachment yet :-)

Ss
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ZenotureAuthor Commented:
Yea, I can't get it to for some reason, regardless, the links work! =)
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ozoCommented:
So how do you get x = 4/3?
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ZenotureAuthor Commented:
taking the derivative of sqrt(1+x^2)/4 + 1-x/5 results in x = 4/3, sorry I should have clarified that better, i thought is was implied where I wrote "3 or 4"
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ZenotureAuthor Commented:
true, but a domain is set regardless, we know based on common sense we would have results greater than or equal to 0, and no greater than or equal to 1, since both given distances are 1 mile each.
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ozoCommented:
Either way, you can reject  x = 4/3 as an answer
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ZenotureAuthor Commented:
So then x = 4/3 is the only possible answer that will work? But that only solves part A, does that mean that there is no reasonable answer for part B?
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ozoCommented:
A reasonable answer for part B would be 0<=x<=1
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ZenotureAuthor Commented:
And since the critical number from the first derivative of that function does not fall within the domain set, we can sat that x = 3/4 is the point at which the time traveled to destination will be the least, correct?
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ZenotureAuthor Commented:
is the only point*
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ozoCommented:
In part A, there are no points within the domain set for which the travel time is less than at  x = 3/4.
Where in the domain set is the travel time the least in part B?
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deightonprogCommented:
how about calling 'x' the angle the boat makes with the vertical, angle  OBP

then with trigonometry

BP = 1 / cos x = sec x

PT = 1 - OP = 1 - tan x

then

time,
t = (1/3) sec x + (1/5) - (1/5) tan x

dt/dx = (1/3) sec x tan x - (1/5) sec^2(x)

need

(1/3) tanx - (1/5) sec x = 0
leads to

sin x = (3/5)

tan(x) = sin x / cos x = (3/5) / sqrt(1 - (3/5)^2) = (3/5)/ sqrt(16/25) = 3/4

so OP works out as 3/4

think you inverted it somewhere



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deightonprogCommented:
sorry, for the second,

sin(x) = (4/5)

tan(x) = (4/5) / [1 - (4/5)^2] = (4/5) / 3/5 = 4 / 3

I've got the same problem as you

----------------------

I've seen what's wrong, the walk from P to T would take 'negative time' in the formula if x>1, but it would take positive time in reality, you need a different formula for x> pi/4 - as ozo said though already.  
General reasoning says that if he rows straight there, then it has to take longer to row to any point beyond P and walk back (as you have correctly reasoned), for the purpose of the question you know that there is no minimum or maximum on 0 to 1, so if you show that rowing straight there is faster than rowing to 0 then walking a mile, you have proved that rowing straight there has to be fastest (and rowing to 0 and walking a mile is slowest)



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BigRatCommented:
>>how about calling 'x' the angle the boat makes with the vertical, angle  OBP

No, I'd just simply call the point at the right angle O, and the distance OP x and solve quite simply.

Isn't it interesting how in nicely constructed puzzles like this how 3,4 and 5 turn up?
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deightonprogCommented:
the educative point of the question is that the solution might not be where f'(x) = 0.  

In fact f'(x) is discontinuous and flips from being negative to positive when the rower heads straight for the target - but if you draw a graph of time taken by landing point, it will have a minimum, but not a minimum where the first derivative is continuous
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deightonprogCommented:
BigRat - it's equally as easy to get the second part wrong using trigonometry - that's what I found.

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