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Problem of the Week #59 - May 13th, 2013

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Chris L T521

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Jan 26, 2012
Thanks again to those who participated in last week's POTW! Here's this week's problem!


Problem: A steel pipe is being carried down a hallway 9 ft wide. At the end of the hallway there is a right-angled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner?


Remember to read the POTW submission guidelines to find out how to submit your answers!
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Chris L T521

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Jan 26, 2012
This week's problem was correctly answered by MarkFL and TheBigBadBen.

Here's MarkFL's solution (which uses Lagrange multipliers):
If we ignore all dimensions of the pipe except the length, this is equivalent to minimizing the sum of the squares of the intercepts of a line passing through the point $(r,s)$ in the first quadrant. Let $a$ and $b$ be the $x$-intercept and $y$-intercept respectively. Thus, the function we wish to minimize is (the objective function):


Now, using the two-intercept form for a line, we find we must have (the constraint):

$\displaystyle \frac{r}{a}+\frac{s}{b}=1$

Using Lagrange multipliers, we find:

$\displaystyle 2a=\lambda\left(-\frac{r}{a^2} \right)$

$\displaystyle 2b=\lambda\left(-\frac{s}{b^2} \right)$

and this implies:

$\displaystyle b=a\left(\frac{s}{r} \right)^{\frac{1}{3}}$

Substituting for $b$ into the constraint, there results:

$\displaystyle \frac{r}{a}+\frac{s}{a\left(\frac{s}{r} \right)^{\frac{1}{3}}}=1$

$\displaystyle a=r^{\frac{1}{3}}\left(r^{\frac{2}{3}}+s^{\frac{2}{3}} \right)$

Hence, we have:

$\displaystyle b=s^{\frac{1}{3}}\left(r^{\frac{2}{3}}+s^{\frac{2}{3}} \right)$

and so we find:

$\displaystyle f_{\min}=f\left(r^{\frac{1}{3}}\left(r^{\frac{2}{3}}+s^{\frac{2}{3}} \right),s^{\frac{1}{3}}\left(r^{\frac{2}{3}}+s^{ \frac{2}{3}} \right) \right)=\left(r^{\frac{2}{3}}+s^{\frac{2}{3}} \right)^3$

Now, we need to take the square root of this since the objective function is the square of the distance we actually wish to minimize. Let $\ell$ be the length of the pipe, and we now have:

$\displaystyle \ell_{\max}=\left(r^{\frac{2}{3}}+s^{\frac{2}{3}} \right)^{\frac{3}{2}}$

Letting $r=6\text{ ft}$ and $s=9\text{ ft}$ we have:

$\displaystyle \ell_{\max}=\left(6^{\frac{2}{3}}+9^{\frac{2}{3}} \right)^{\frac{3}{2}}\text{ ft}\approx21.07044713766\text{ ft}$

Here's TheBigBadBen's solution:
I begin with intuitive assumptions:

1: The straight pipe can be made arbitrarily thin
2: An optimal way to manoever the pipe is to keep it flush against the corner of the turn, rotating along the 90 degree bend.
3: The height of the hallway doesn't factor in (i.e. is sufficiently low, I guess).

With that in mind, a pipe can pass through if at all points in this rotation it does not hit the wall.

Long story short, we frame the problem as follows:

"consider a line of the form
(y - 9) + m(x - 6) = 0
for m >= 0. Find the minimum length of the line passing through the first quadrant"

This amounts to a routine minimization problem. We note that for such a line, the axis intersections are given by
(9/m + 6, 0) and (0, 9 + 6m)

The desired length d then satisfies

(9/m+6)^2 + (9 + 6m)^2 = d^2

expanding this, we get

d^2 = 36 m^2 + 108 m + 117 + 108/m + 81/m^2

Now, we find the critical points:

(d/dm)d^2 = 72 m + 108 - 108/m^2 - 162/m^3 = 0

Rejecting the case that m = 0, we multiply by m^3 and factorize to get

18(2m + 3)(2m^3 - 3) = 0

The only root in the desired domain is
m = (3/2)^(1/3)

Plugging in this solution yields the answer
d ~= 21.07
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