The obvious answer is that the bug crawls straight down a vertical edge (3 inches), and then travels the diagonal of the bottom rectangular side. Pythagoras tells us this diagonal has length (5² + 12²)^1/2 = 13 inches, so the total travel distance of 16 inches. This analysis, however, is wrong (and why the puzzle is so devilishly clever). The actual path the bug should take may be more easily seen by mentally taking off the top of the box, cutting along the vertical edges, and folding the sides down onto the same plane as the bottom. Then one can see a straight-line path from the one corner to its opposite as the hypotenuse of a right triangle with sides of 5 inches and 12+3=15 inches. This hypotenuse becomes the optimal path for the bug after the sides are folded back up; its length being (5² + 15²)^1/2 = 250^1/2 ~= 15.8 inches.

Here is a variation on this puzzle that has an even more obvious wrong answer; I found it in a 1962 edition of "Mathematics for Pleasure," by Oswald Jacoby and William H. Benson: Instead of a shoebox, we have a room that is 12 ft. wide, 12 ft. high, and 30 ft. long. The bug, starting on one square wall 1 ft. from the ceiling and 6 ft. from either adjoining wall, wants to crawl to the opposite wall, 1 ft. from the floor and 6 ft. from either adjoining wall. How long is the shortest possible path?

(alternative solution)