You have a corridor which has an L-shape in it. The corridor looks like this:

where a and b are the widths of the sections of the corridor. The question is to find the longest pipe that can be carried down this corridor. The word pipe here just means something long and with essentially 0 thickness. There is a huge simplification which is being assumed here, which is that the corridor is only 2 dimensional. Of course in a 3 dimensional corridor we have a lot more room to manoeuvre.

Let’s think about a pipe going round the corner. The longest pipe that can go through is the length of the shortest gap that it has to go through. So let’s think of a pipe at a particular angle with respect to the corner:

The pipe here is the blue line and is the angle that it makes with respect to the corner. What we need to find is the length of the longest pipe that can fit in for a given . We can see that the length of this pipe is:

Now we know that the pipe will need to go right around the corner which means that it has to go through all in . We use the open interval as we can see that the function is divergent at 0 and . What is the minimum as a function of ? Because the function is divergent and positive at the endpoints we don’t need to check whether these are minima. Thus we need to see if there are any turning points in the function:

equating to zero, this can be simplified to give:

and thus:

as is 1:1 there is a unique critical point, and because we know that the function is continuous on the open interval, and that it is divergent and positive at the endpoints, this must be where the minimum is. Thus, we can calculate the value of at this point:

This last part takes some manipulation.

Therefore, this is the longest pipe that can go around the corner, as it is the shortest distance that it must go through in its journey.

Anita CampbellNovember 3, 2018 at 7:47 pmIt was good to point out that this is a 2D problem – maybe the pipe has sand in that we don’t want to spill so it can’t be tilted.

Looks like a small typo in \theta=\arctan\left(\sqrt{3}\frac{b}{a}\right). It is correct when substituted into l(theta).