Answer to

old chestnut: carrying a ladder around a corner:

At the maximum length, at one point during the journey, the two ends of the ladder are rubbing against the two outer walls, and the side of the ladder is rubbing against the inner corner. You can draw a series of lines like this, which represent the longest length ladders which can be placed at various angles in this corner. We want to find the length of the shortest one of these lines -- that will be the longest ladder which can go all the way through the corner.

If the hallways are of the same width, the puzzle is much easier. The critical position is, by symmetry, that in which the ladder is at a 45 degree angle with respect to the walls, and thus the length of the ladder protruding into each section of hallway is sqrt(2) times the hallway width, so the whole ladder is 2*sqrt(2) times the hallway width.

The general problem requires a more mathematical solution using derivatives. Consider the set of lines from the first paragraph, all touching the inner corner and running from outer wall to outer wall, but of varying lengths. Determine the length of these lines with respect to angle and *A* and *B*, and minimize the length with respect to angle.

Let *x* be the acute angle between the ladder and the wall on the *B* side. The similar angle on the A side is 90 degrees minus *x*. As a result, the length of the ladder on the *B* side is *B*/sin(*x*) and the length on the *A* side is *A*/cos(*x*).
Total length *L* = *B*/sin(*x*) + *A*/cos(*x*).

Now set d*L*/d*x* = 0.

So d/d*x*{*B*/sin(*x*) + *A*/cos(*x*)} = 0

d/d*x*{*B* csc(*x*) + *A* sec(*x*)} = 0

- *B* csc(*x*)cot(*x*) + *A* sec(*x*)tan(*x*) = 0

*B* cos(*x*)/sin^{2}(*x*) = *A* sin(*x*)/cos^{2}(*x*)

*B* cos^{3}(*x*) = *A* sin^{3}(*x*)

*B*/*A* = tan^{3}(*x*)

(*B*/*A*)^{1/3} = tan(*x*)

so *x* is the inverse tangent of the cube root of *B*/*A*.

You can plug this into the original formula, but it's possible to simplify as follows:

Since 1 + tan^{2}(*x*) = sec^{2}(*x*) = 1/cos^{2}(*x*), we can write

cos(*x*) = 1/sqrt(1+tan^{2}(*x*)), or

cos(*x*) = 1/sqrt(1+(*B*/*A*)^{2/3})

so *A*/cos(*x*) = *A* sqrt(1+(*B*/*A*)^{2/3})

Now sin(*x*) = tan(*x*)cos(*x*), so

sin(*x*) = (*B*/*A*)^{1/3}/sqrt(1+(*B*/*A*)^{2/3})

which gives *B*/sin(*x*) = *B* sqrt(1+(*B*/*A*)^{2/3}) / (*B*/*A*)^{1/3}

So *L* = *A*/cos(*x*) + *B*/sin(*x*) = (*A* + *B*/(*B*/*A*)^{1/3}) sqrt(1+(*B*/*A*)^{2/3})

Divide out an *A*^{1/3} from the first part of this expression:

*L* = (*A*^{2/3} + *B*^{2/3}) *A*^{1/3} sqrt(1+(*B*/*A*)^{2/3})

= (*A*^{2/3} + *B*^{2/3}) sqrt(*A*^{2/3}+*B*^{2/3})

= (*A*^{2/3} + *B*^{2/3})^{3/2}