Math problem for the geeks!

[weirdichi]

Text
What is the longest object that can be carried through this corner without distoriting the hall or the object?

I'm thinking the shortest distance is a 45 degree angle from either wall of the hall. My geometry is rusty, so I don't know how to compute the extension from the initial measurements on the walls. Any hints?

 

[ucdbiendog]

its not 45 degrees because the halls are not te same size. too lazy to figure it out myself though

edit: to figure it out though, construct an equation that gives the length in terms of theta (som angle with one of the walls) and then take the derivative of that equation and set it to zero and solve for theta

 

[DOSfan]

First, it really depends on the width of the object.

Baring that little detail, draw triangles. That should help you figure out the remaining numbers.

That should be enough of a hint for you and your friend.

 

[weirdichi]

Originally posted by: ucdbiendog
its not 45 degrees because the halls are not te same size. too lazy to figure it out myself though

edit: to figure it out though, construct an equation that gives the length in terms of theta (som angle with one of the walls) and then take the derivative of that equation and set it to zero and solve for theta

True.. they won't be 45 degrees because of the different widths. I'm lost now then.

 

[ucdbiendog]

Originally posted by: DOSfan
First, it really depends on the width of the object.

Baring that little detail, draw triangles. That should help you figure out the remaining numbers.

That should be enough of a hint for you and your friend.

width does not apear to be an issue looking at the picture. At any rate, my method would still apply, it would just be a more complicated function

 

[weirdichi]

Originally posted by: DOSfan
First, it really depends on the width of the object.

Baring that little detail, draw triangles. That should help you figure out the remaining numbers.

That should be enough of a hint for you and your friend.

I think it's just a matter of length, disregarding width. Also, i think it's just on a 2D plane, as you can't hold one end higher than the other end to "shorten" the overall length of the object.

 

[ucdbiendog]

Originally posted by: weirdichi

Originally posted by: ucdbiendog
its not 45 degrees because the halls are not te same size. too lazy to figure it out myself though

edit: to figure it out though, construct an equation that gives the length in terms of theta (som angle with one of the walls) and then take the derivative of that equation and set it to zero and solve for theta

True.. they won't be 45 degrees because of the different widths. I'm lost now then.

ok say theta is the angle btw the object and that bottom wall. using the geometry, you can find the length of the object for an arbitrary theta. so:
L=f(theta)

then set dL/d(theta) = 0 and solve for theta. if you havent had basic calc, i dont know how else to explain it, thats the easiest way to do it i think

 

[ucdbiendog]

just remember that sin(theta) = opposite/hypotenuse, and cos(theta) = adjacent/hypotenuse, and you can solve for the two hypotenuses (hypoteni?)

 

[weirdichi]

Thanks for the hints. I'll make her send you a pic.

 

[chuckywang]

The answer is 6*sqrt(1+cubert(16/9))+8*sqrt(1+cubert(9/16))

cubert = cube root.

 

[eLiu]

This is kind of a 'classic' optimization problem (or that's what it's called in my old calc book). So you need to turn this into an eqn of one variable so you can take the deriv & find the max. You do this by finding the equation for length of bar that will fit around the corner giving some angle theta. Draw a triangle, it isn't hard.

Now I'm assuming the hallways are of the same *height*. It's more complicated if they aren't.

 

[kponds]

14

 

[WiseOldDude]

tell your lazy ass friend to figure it our for himself. if he happens to graduate from school and get a job at something higher than working at mcdonalds, he will be required to do his job all by himself.

 

[Triumph]

I figured it out with straight geometry, recognizing that the longest length will be achieved when the line is parallel to the hypotenuse made by the 6 and 8. My answer is -1.2684.

 

[chuckywang]

Originally posted by: Triumph
I figured it out with straight geometry, recognizing that the longest length will be achieved when the line is parallel to the hypotenuse made by the 6 and 8. My answer is -1.2684.

Your answer is negative? Doesn't sound right.

 

[Tuffrabbit]

The lenght of the diagonal line calculates out to 19.79898987......