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Puzzle: another boat

May 23, 2007 pm31 3:32 pm

The previous question seemed so easy. Let’s modify it, just a bit:

On a lake, with a very straight shore, you are 4 miles out in a small boat. You can row at 4mph, but once you reach land, you need to get to a friend who is 12 miles up the shore. You have your skateboard with you , and can ride at 12 mph. (in other words, the perpendicular boat-shore and the shore itself create a right triangle). What’s the least amount of time it would take you to reach your friend?

3 Comments leave one →
  1. Clueless permalink
    May 23, 2007 pm31 5:43 pm 5:43 pm

    Well,

    If it takes you more than about 3.43 minutes to calculate the right point to reach on the shore with the boat, it is probably better to just follow the legs of the right triangle.

    Clueless.

  2. rdt permalink
    May 24, 2007 am31 12:52 am 12:52 am

    All my calculus instincts, and all my gaming the test instincts tell me to guess that taking the legs of the right triangle is the way to go: you spend equal amounts of time rowing and skate boarding, you minimize the time you spend rowing, and the numbers work out nicely when you do the calculation…

    But the real way to do it is minimizing:

    L/12 + ([4^2 + (12-L)^2]^(1/2))/4

    — Rachel (who’s blanking on Latex syntax just now…)

  3. May 26, 2007 pm31 7:51 pm 7:51 pm

    I think I like clueless’ answer. Of course, we could go on to ask, boat B miles out, speed b, landspeed s, distance S, for what values of B/S, b and s do we follow the legs? the hypotenuse?

    But that feels just like an exercise, and fairly tedious at that. I may try it privately, but it’s not for this blog.

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