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

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#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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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?

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

#### Chris L T521

##### Well-known member
Staff member
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):

$f(a,b)=a^2+b^2$

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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