A spider is in a rectangular warehouse measuring 40 x 10 x meters. The spider is on the 10 x 10-meter wall, five meters from the sides and one meter above the ground. the proverbial fly is on the opposite wall, five meters from the sides and one meter below the ceiling. what is the shortest route for the spider to walk to the fly?

Question

anonymous · Answer

I can't get this one :(

anonymous · Answer

Posting another one..

anonymous · Answer

40 x 10 x ?

anonymous · Answer

must be 40x10x10 according to next sentence

anonymous · Answer

I got him going under 50m, not sure if its the shortest distance.  He crawls down the 1m of the 10x10 wall at a 45º angle, so he goes sqrt(2) m.  Then goes to the 40x10 wall at another 45º.  He's already gone 1m toward the 40x10 wall.  So going to the 40x10 wall he has to go 4sqrt(2) m, for a total of 5sqrt(2) mso far.  Then he still has 36 feet of the 40x10 wall to traverse and 5 m of the opposite 10x10 wall horizontally and 9 m vertically.  So he has to go sqrt(41^2+9^2) more m, for a total of sqrt(1762) + 5sqrt(2) m, around 49 m

directrix · Answer

The Spider and the Fly problem is one of the classics. Several variations of the problem exist. Here's one: The Spider and the Fly A spider, in the top-left-front corner of a 10 x 10 x 10 foot room, sees a big fat fly in the bottom-right-back corner. Describe the shortest path, and the length of the path, that the spider can crawl to get the fly. That's crawl, not jump, or fly. Solution: Think of unfolding the room and seeing the ceiling and right wall as a 10 x 20 domino. The shortest path is a straight line; the diagonal of this domino, which is -/(10^2 + 20^2) = -/500 = 10-/5 ft , or about 22.36 feet. Answer: 10-/5 ft , which is 22.36 ft to nearest hundredth. http://www.dansmath.com/probofwk/probar02.html#anchor2428389