1. ## Maximum possible length?

Hello.

This seems pretty fundamental to me, however I was not able to find information about this particular problem on the net, so I'm writing here seeking your knowledge

Let's say we have two connected perpendicular "tunnels" like in this picture:

Widths are x and y respectively. Question is, what is the maximum length of an indefinitely narrow object, which is able to move from one end to another?

Sorry for possible awkward presentation, English is obviously not my native language.

2. Are there any restrictions on the angle the length sits within the tunnel?

3. Originally Posted by pickslides
Are there any restrictions on the angle the length sits within the tunnel?
There are no such restrictions. On the other hand, there is no fixed angle as well
I mean, imagine black lines being "walls" and red line being a narrow "car" (couldn't think of a less lame comparison) - "car" will eventually have to make a nearly 90 degrees turn, but I do not know which particular moment will be the most complicated for the "car" to fit within the "walls"...

4. Originally Posted by Pranas
Hello.

This seems pretty fundamental to me, however I was not able to find information about this particular problem on the net, so I'm writing here seeking your knowledge

Let's say we have two connected perpendicular "tunnels" like in this picture:

Widths are x and y respectively. Question is, what is the maximum length of an indefinitely narrow object, which is able to move from one end to another?

Sorry for possible awkward presentation, English is obviously not my native language.
[IMG]19976[/IMG]

Here is my 2 cents

From my diagram you can see that you are trying to maximize the quantity

$\displaystyle d_1+d_2=\frac{y}{\sin(\alpha)}+\frac{x}{\cos(\alph a)}=M(\alpha)$

$\displaystyle M'(\alpha)=-\frac{y\cos(\alpha)}{\sin^2(\alpha)}+\frac{x\sin(\ alpha)}{\cos^2(\alpha)}$

I used a CAS to solve for $\displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}} \right)$

Then just plug this into $M(\alpha)$

This gives $\displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra c{3}{2}}$

5. Originally Posted by TheEmptySet
[IMG]19976[/IMG]

Here is my 2 cents

From my diagram you can see that you are trying to maximize the quantity

$\displaystyle d_1+d_2=\frac{y}{\sin(\alpha)}+\frac{x}{\cos(\alph a)}=M(\alpha)$

$\displaystyle M'(\alpha)=-\frac{y\cos(\alpha)}{\sin^2(\alpha)}+\frac{x\sin(\ alpha)}{\cos^2(\alpha)}$

I used a CAS to solve for $\displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}} \right)$

Then just plug this into $M(\alpha)$

This gives $\displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra c{3}{2}}$
Thank you very much.

Of course we are minimizing (you wrote maximizing) the quantity. Nevertheless, your method is very graceful and correct, while my initial idea using coordinates was way too complicated.

I am able to approach
$\displaystyle \alpha_m_i_n=\arctan\left( \sqrt[3]{\frac{y}{x}} \right)$
then
$\displaystyle M(\alpha_m_i_n) = y \cdot \sqrt{1 + (\frac{x}{y})^{\frac{2}{3}}} + x \cdot \sqrt{1 + (\frac{y}{x})^{\frac{2}{3}}}$
as long as it seems the same as
$\displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra c{3}{2}}$
I am not quite able to reach this wonderful expression. Could someone please give me a hint?

6. Originally Posted by Pranas
Thank you very much.

Of course we are minimizing (you wrote maximizing) the quantity. Nevertheless, your method is very graceful and correct, while my initial idea using coordinates was way too complicated.

I am able to approach
$\displaystyle \alpha_m_i_n=\arctan\left( \sqrt[3]{\frac{y}{x}} \right)$
then
$\displaystyle M(\alpha_m_i_n) = y \cdot \sqrt{1 + (\frac{x}{y})^{\frac{2}{3}}} + x \cdot \sqrt{1 + (\frac{y}{x})^{\frac{2}{3}}}$
as long as it seems the same as
$\displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra c{3}{2}}$
I am not quite able to reach this wonderful expression. Could someone please give me a hint?
First notice that

$\displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}}\right) \iff tan(\alpha)=\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}} }$

Since $\displaystyle 0 < \alpha \le \frac{\pi}{2}$ it must be in the first quadrant. Use the Pythagorean theorem to find the hypotenuse of the triangle. see diagram

Using this we get that

$\displaystyle \sin(\alpha)=\frac{y^{\frac{1}{3}}}{\sqrt{x^{\frac {2}{3}}+y^{\frac{2}{3}}}} \text{ and } \cos(\alpha)=\frac{x^{\frac{1}{3}}}{\sqrt{x^{\frac {2}{3}}+y^{\frac{2}{3}}}}$

Now just plug this in to get

$\displaystyle M(\alpha)=\frac{y}{\frac{y^{\frac{1}{3}}}{\sqrt{x^ {\frac{2}{3}}+y^{\frac{2}{3}}}}}+\frac{x}{\frac{x^ {\frac{1}{3}}}{\sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3 }}}}}=y^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^{\fra c{2}{3}}}}+x^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^ {\frac{2}{3}}}}=$

$\displaystyle \sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3}}}}\left( x^{\frac{2}{3}}+y^{\frac{2}{3}}}\right)=\left(x^{\ frac{2}{3}}+y^{\frac{2}{3}}}\right)^{\frac{3}{2}}$

7. Originally Posted by TheEmptySet
First notice that

$\displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}}\right) \iff tan(\alpha)=\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}} }$

Since $\displaystyle 0 < \alpha \le \frac{\pi}{2}$ it must be in the first quadrant. Use the Pythagorean theorem to find the hypotenuse of the triangle. see diagram

Using this we get that

$\displaystyle \sin(\alpha)=\frac{y^{\frac{1}{3}}}{\sqrt{x^{\frac {2}{3}}+y^{\frac{2}{3}}}} \text{ and } \cos(\alpha)=\frac{x^{\frac{1}{3}}}{\sqrt{x^{\frac {2}{3}}+y^{\frac{2}{3}}}}$

Now just plug this in to get

$\displaystyle M(\alpha)=\frac{y}{\frac{y^{\frac{1}{3}}}{\sqrt{x^ {\frac{2}{3}}+y^{\frac{2}{3}}}}}+\frac{x}{\frac{x^ {\frac{1}{3}}}{\sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3 }}}}}=y^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^{\fra c{2}{3}}}}+x^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^ {\frac{2}{3}}}}=$

$\displaystyle \sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3}}}}\left( x^{\frac{2}{3}}+y^{\frac{2}{3}}}\right)=\left(x^{\ frac{2}{3}}+y^{\frac{2}{3}}}\right)^{\frac{3}{2}}$
Indeed.
Thanks again.
That's a shame I was not able to figure this out myself, I need some more practicing, I guess...