An easier version of this puzzle is here.

A large panel needs to be moved through a corridor, the panel is tall as the corridor. The corridor is A feet wide before a right angle turn, after the turn, it is B feet wide. What is the maximum length of the panel that can pass through this corner.

Overhead view of the hallway:

+------------+---
|           /   |
|          /    |B ft
|         /     |
|        /+------
|       / |
|      /  |
|     /   |
|    /    |
|   /     |
|  /      |
| /       |
|/        |
+<-A ft-->|

As the panel "swings" around the corner, the point at which it is "most constrained" by the walls of the corridor will be when the panel makes a 45 degree angle with the corner.

At this point, the panel would be the sum of the hypotenuses of two isoceles triangles one with leg a and one with leg b.

This means it has length = [ √(2a²) + √(2b²) ]
__________________________

One may question whether this minimization occurs at a 45 degree angle. I think it is self evident that this happens if one considers the motion of the end points along the wall as the panel is rotated at the corner.

However, (because this is posted in Calculus) if one wishes to consider it, then one can differentiate the maximum length of the panel as a function of the angle it makes with the corner, take the derivative, set equal to zero (to find the minimum), and evaluate at that point.

I leave this as an exercise to the reader! :-)

Posted by SilverKnight on 2004-05-26 14:28:31