Visualization tip: Consider the possible nets of the rectangular prism that makes up the room, and find what the shortest path on one of them would look like.
Another way of thinking about the visualization is to make a model of the surface that can fold up to form it. The shortest path (which will be a diagonal if not a straight line) on any of those is the answer.
I ended up with the right answer by finding the minimum of the equation
f(x,y) = 2(sqrt(1+x^2) + sqrt((6-x)^2 + y^2) + sqrt((15-y)^2 + 36))
I should have thought to unravel the room! Errr.
Some limits can be placed on the number of possible paths by considering some general properties of an optimal solution:
1. Assuming the start point is A, and the path leaves the start side at point B, the optimal path will take a straight line from A to B. Proof: if it were not straight, it could be replaced by a straight line which would shorten the path, contradicting the assumption the path was optimum.
2. Once the optimal path leaves a side, it will not return to that side. Proof. Suppose the path leaves the side at point B, then later enters at C, and then leaves again at D. The route from B to C could be replaced with a straight line from C to D, which would shorten the path, again contradicting the assumption that the path was optimum.
This cuts down the number of possible path templates greatly, since a given surface can only appear once in the optimal path, and only contain a single straight line segment.
I expect that a little more reasoning can deduce some more limits on possible optimal paths, to get it down to just two or three possible equations to maximize.
sqrt(1+x^2) + sqrt((6-x)^2+y^2)+sqrt(12^2+z^2)+
sqrt((30-z-y)^2+u^2)+sqrt((6-u)^2+1)
40 at {x -> 0.75, y -> 7, z -> 16, u -> 5.25}
your equation: http://www.wolframalpha.com/input/?i=minimize+2%28sqrt%281%2...
Ant jumps off the wall and onto the floor (no walking necessary, and easily survivable), walk straight across the 30 foot floor and 1 foot up the wall.
It's a flying ant.