Max/Min Hallway Problem

May 2010
2
0
Okay: A pipe of negligible diameter is to be carried horizontally around a corner from a hallway 8 feet wide into a hallway 4 feet wide.
a) what is the maximum length that the pipe can have? and
b) if we set the pipe's diameter at 1 inch, how does this affect the problem?

This is beyond me. I set up a diagram and think I'm supposed to be finding the max L, when L= x+y, where x is the length of the pipe in the 8 foot hallway and y is the length of the pipe in the 4 foot hallway. After this, i don't know where to proceed. I think it needs trig or similar triangles perhaps. Please help? Thanks so much... :)
 
Last edited:
Apr 2010
384
153
Canada
Okay: A pipe of negligible diameter is to be carried horizontally around a corner from a hallway 8 feet wide into a hallway 4 feet wide.
a) what is the maximum length that the pipe can have? and
b) if we set the pipe's diameter at 1 inch, how does this affect the problem?

This is beyond me. I set up a diagram and think I'm supposed to be finding the max L, when L= x+y, where x is the length of the pipe in the 8 foot hallway and y is the length of the pipe in the 4 foot hallway. After this, i don't know where to proceed. I think it needs trig or similar triangles perhaps. Please help? Thanks so much... :)
For part a I will attempt to construct a diagram.
..........what the hell, my diagram got ****ed up when I posted...so we will have to explain

The longest pipe will have the length equal to the minimum of \(\displaystyle L= x+y \) where x is the distance from the top of the pipe in the first hallway to the edge where the hallways meet, and y is the distance from the edge where the hallways meet to the point where the pipe hits the wall. The angle from the horizontal of the second hallway to the line defined by y is going to be theta.

Thus,

\(\displaystyle x = \frac{8}{cos \theta } \) and \(\displaystyle y = \frac{4}{sin \theta } \)

\(\displaystyle L = L( \theta ) = \frac{8}{cos \theta } + \frac{4}{sin \theta } \) where \(\displaystyle 0 < \theta < \frac{ \pi }{2} \)

If \(\displaystyle L`( \theta ) = 0 \) then,

\(\displaystyle \frac { 8 sin \theta }{cos^2 \theta} - \frac{4cos \theta }{sin^2 \theta } = 0 \)

\(\displaystyle \frac{ 8 sin^3 \theta - 4 cos^3 \theta }{ cos^2 \theta sin^2 \theta } \)

\(\displaystyle 8sin^3 \theta - 4cos^3 \theta = 0 \)

\(\displaystyle tan^3 \theta = \frac{4}{8} \)

\(\displaystyle tan \theta = ( \frac{4}{8} )^{ \frac{1}{3} } \)

\(\displaystyle L( \theta ) --> \infty \) as \(\displaystyle \theta --> 0+ \) or \(\displaystyle \theta --> \frac{ \pi }{2} - \)

Thus the minimum must occur at \(\displaystyle \theta = tan^{-1} ( \frac{1}{2} )^{ \frac{1}{3} } \)

Let us go back to,

\(\displaystyle tan \theta = ( \frac{4}{8} )^{ \frac{1}{3} } \) we can see that (by constructing the triangle and using pythag),

\(\displaystyle cos \theta = \frac{8^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } } \)

and

\(\displaystyle sin \theta = \frac{4^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } } \)

Hence,

\(\displaystyle L( \theta ) = \frac{8}{ \frac{8^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } } } + \frac{4}{ \frac{4^{ \frac{1}{3} } }{ \sqrt{ 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } } } } \)

\(\displaystyle L = ( 8^{ \frac{2}{3} } + 4^{ \frac{2}{3} } )^{ \frac{3}{2} } \)
 
May 2010
2
0
part b

thanks. i ended up figuring it out by similar triangles as well....
Now for part b. If the pipe is given a diameter of 1 inch, how will it affect the problem?

Now i know it will shorten the pipe because the point where the pipe theoretically touches at its longest length will be... .5 inch away? Not sure how to exactly expand on that.
 
Apr 2010
384
153
Canada
thanks. i ended up figuring it out by similar triangles as well....
Now for part b. If the pipe is given a diameter of 1 inch, how will it affect the problem?

Now i know it will shorten the pipe because the point where the pipe theoretically touches at its longest length will be... .5 inch away? Not sure how to exactly expand on that.
Draw the original diagram and superimpose it with the new one. What do you notice? Well, our stick which was originally as thin as our pencil marking now has an edge that hits both the walls and the edge where the hallways intersect. This forces us to have a greater angle of theta (as defined in my previous post).

And note that our original analysis is no longer valid because of the position of our new beam, the part of the beam that touches both walls sits an inch off of the edge where the hallways intersect. Thus, similar triangles are out the window as well as my approach.

How do we set this up then? There are a number of ways but off the top of my head I would analyze the side that hits both the walls but hangs off the edge by the diameter of the beam. Re-define your theta and do an analysis of triangles.