Re: trip problem

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Posted by Brad Paul on November 05, 2002 at 07:52:56:

In Reply to: Re: trip problem posted by RD on November 04, 2002 at 17:29:08:

: : : This is apparently from an old walk and bike puzzle:

: : : 2 girls and a guy want to travel 200 miles on a straight path.
: : : All 3 are together, and have one bike.
: : : The girls walk at 2mph, bike at 24mph.
: : : The guy walks at 9mph, bikes at 44mph.
: : : Only one can use the bike at a given time.
: : : What is the minimum time in which they can all
: : : get to the 200 mile point, and how do they do it?

: : : I've been trying this for a while but can't get anywhere.
: : : Can someone help, please.
: : I would just put the two girls in the canoe and bike the 200 miles.
: : http://bradpaul.bdp-optics.com/trips/mis/mis_pic.html

: I'm not interested in your web page, Mr P.
: That puzzle is a mathematical problem and I need help with it.
: Quit making fun of me.
Sorry my intent was not to make fun of you.

Let the distance be d.

The speed of the girls on foot be v_gf and the speed of the
girls on the bike be v_gb.

The two speeds for the boy are v_bf and v_bb.

The fastest way for the girls would be to both use the bike half
way. If this is faster than the boy can go by foot we need to have the
boy use the bike some. To do this right we need to parametrize the
question such that we can us our knowledge of calculus. Imagen if we
let girl (A) bike from 0 to f d/2 where f is between 0 and 1. Then the boy
after walking from 0 to f d/2 bikes from f d/2 to d-(f d/2) where he leaves
the bike and starts walking. Girl B will picks up the bike and bikes
from d-(f d/2) to d. The time it take both of the girls is the same.

How long does it take a girl to do the distance?

t_g=(walking_distance)/v_gf+(biking_distance)/v_gb
t_g=(d-(f d/2))/v_gf+(f d/2)/v_gb

How long does it take the boy to do the distance?

t_b=2(f d/2)/v_bf+(d-(f d/2)-f d/2)/v_bb
t_b=f d/v_bf+(d(1-f))/v_bb

We need to find the min of max(t_g(f),t_b(f))
Given f is between 0 and 1.

If you put in the numbers and plot these to function you will find is
is fastest to have f=1. Which means the boy never bikes. Which brings
me back to my original idea of putting the two girls in the canoe.

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No, that's not it, Mr P RD 10:56:32 11/05/02 (0)

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: : : : This is apparently from an old walk and bike puzzle: : : : : 2 girls and a guy want to travel 200 miles on a straight path. : : : : All 3 are together, and have one bike. : : : : The girls walk at 2mph, bike at 24mph. : : : : The guy walks at 9mph, bikes at 44mph. : : : : Only one can use the bike at a given time. : : : : What is the minimum time in which they can all : : : : get to the 200 mile point, and how do they do it? : : : : I've been trying this for a while but can't get anywhere. : : : : Can someone help, please. : : : I would just put the two girls in the canoe and bike the 200 miles. : : : http://bradpaul.bdp-optics.com/trips/mis/mis_pic.html : : I'm not interested in your web page, Mr P. : : That puzzle is a mathematical problem and I need help with it. : : Quit making fun of me. : Sorry my intent was not to make fun of you. : Let the distance be d. : The speed of the girls on foot be vgf and the speed of the : girls on the bike be vgb. : The two speeds for the boy are vbf and vbb. : The fastest way for the girls would be to both use the bike half : way. If this is faster than the boy can go by foot we need to have the : boy use the bike some. To do this right we need to parametrize the : question such that we can us our knowledge of calculus. Imagen if we : let girl (A) bike from 0 to f d/2 where f is between 0 and 1. Then the boy : after walking from 0 to f d/2 bikes from f d/2 to d-(f d/2) where he leaves : the bike and starts walking. Girl B will picks up the bike and bikes : from d-(f d/2) to d. The time it take both of the girls is the same. : How long does it take a girl to do the distance? : tg=(walking_distance)/vgf+(biking_distance)/vgb : tg=(d-(f d/2))/vgf+(f d/2)/vgb : : How long does it take the boy to do the distance? : tb=2(f d/2)/vbf+(d-(f d/2)-f d/2)/vbb : tb=f d/vbf+(d(1-f))/vbb : We need to find the min of max(tg(f),tb(f)) : Given f is between 0 and 1. : If you put in the numbers and plot these to function you will find is : is fastest to have f=1. Which means the boy never bikes. Which brings : me back to my original idea of putting the two girls in the canoe.

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