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,

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

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