Check out some similar questions!

galactus · Jan 12, 2007, 06:05 AM

This site doesn't have enough traffic. So here's a max/min for those who would like to try. Maybe you've seen something similar before.

"A pipe of negligible diameter is to be carried horizontally around a corner from a hallway 8 ft wide into a hallway 4 ft wide. What is the maximum length the pipe can have?".

Realistically, the diameter would have to be included, but then the problem becomes more complex. There is a general formula for this type of problem if anyone wants to try and derive it by using random variables.

Curlyben · Jan 12, 2007, 06:10 AM

So if the pipe has neligable diameter why not just bend it around the corner or even better coil it up and throw it over your shoulder ;)

galactus · Jan 12, 2007, 06:13 AM

That's no fun:D

Capuchin · Jan 12, 2007, 01:06 PM

Infinite rigidity? :)

galactus · Jan 12, 2007, 01:25 PM

Come on folks. A little calculus. Like a lot of problems, it's not meant to be actual and realistic. Just a fun calc problem.

asterisk_man · Jan 12, 2007, 01:55 PM

Call the angle between the pipe and the wall of the 8' wide hall $\theta_8$ .
Call the angle between the pipe and the wall of the 4' wide hall $\theta_4$ .
Call the piece of pipe from the corner into the 8' wide hall $L_8$
Call the piece of pipe from the corner into the 4' wide hall $L_4$
Call the total pipe length $L_{total}$

$ \sin \theta_8 = \frac 8 {L_8}\\ L_8 = \frac 8 {\sin \theta_8} $
similarly:
$L_4 = \frac 4 {\sin \theta_4}$

So:
$L_{total}=L_8 + L_4=\frac 8 {\sin \theta_8} + \frac 4 {\sin \theta_4}$

$\theta_8 + \frac \pi 2 + \theta_4 = \pi\\ \theta_8=\frac \pi 2 - \theta_4 $

therefore

$L_{total}=\frac 8 {\sin \left( \frac \pi 2 - \theta_4 \right) } + \frac 4 {\sin \theta_4}$

Ha! I'm out of time!

I think I'd take the derivative, find the zero in the range $\left[0,\frac \pi 2\right]$ and then plug that into the $L_{total}$ equation. Who wants to try that?

Although I guess I'm not 100% sure that I'm going about this the right way. But someone will feel free to tell me I'm wrong if I am :)

Nosnosna · Jan 12, 2007, 02:09 PM

All right, I'll take a stab at it.

The method is simple: Determine a generalized formula for the length of a line intersecting both outside walls and the point at the corner. This is simply a matter of trig.

There are two components of the length: Distance from the corner to the wall of the 8' hall, and distance from the corner to the wall of the 4' hall Both of these are defined by the angle the line forms at the corner, theta. Theta is arbitrarily defined here as the angle between the line and the interior wall of the 8' hall.

The length of the two segments of the line are given by
sec(theta) * 4
csc(theta) * 8

Thus, the generalized length of the line is:
l = 4 sec(theta) + 8 csc(theta)

Take the derivative of this to get:

l' = 4 sec(theta)tan(theta) - 8 cos(theta)cot(theta)

Set l' = 0 and solve for theta to get the angle with the shortest associated line:

theta ~= 51.5 degrees

And then simply plug that theta into the function for l to get the maximum length.

l = 4 sec(51.5) + 8 csc(51.5)
~= 16.65 feet.

Disclaimer: I'm way out of practice on my math, so while I'm confident in the method, I'm not confident in the numbers :)

Nosnosna · Jan 12, 2007, 02:33 PM

For the even more general case of a pipe of diameter 'd' being passed around the corner, the method is the same, except that instead of having the line intersect the corner, you have the line tangent to a circle of radius d centered on the corner.

galactus · Jan 12, 2007, 02:51 PM

Nosnosna's answer is very good. I like it using trig. My method was similar triangles instead.

The max length of the pipe will be the smallest value of L=x+y

Break the pipe up into 2 sections of x and y. y being the length of the section through the 8 foot hallway and x be the length thorugh the 4 foot hallway(I should've labelled them in the diagram).

Then we have:

$\frac{y}{8}=\frac{x}{\sqrt{x^{2}-16}}, \;\ y=\frac{8x}{\sqrt{x^{2}-16}}$

$L=x+\frac{8x}{\sqrt{x^{2}-16}} \;\ for x>4$

$\frac{dL}{dx}=1-\frac{128}{(x^{2}-16)^{\frac{3}{2}}}$

$\frac{dL}{dx}=0 \;\ when \;\ (x^{2}-16)^{\frac{3}{2}}=128$

$x^{2}-16=128^{\frac{2}{3}}=16(2^{\frac{2}{3}})$

$x^{2}=16(1+2^{\frac{2}{3}})$

$x=4(1+2^{\frac{2}{3}})^{\frac{1}{2}}$ ... [1]

We can check the 2nd derivative and all that, but suffice it to say that L is smallest when x equals [1]

For this value of x, $L=4(1+2^{\frac{2}{3}})^{\frac{3}{2}}\approx{16.65 \;\ feet$

galactus · Jan 13, 2007, 06:01 AM

BTW, for those who are curious, there is a general formula for this type of problem.

$(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}$

Enter a=8 and b=4 and you'll get 16.65.