How do you solve plumber's pipe carried around the corner?

You have already many excellent correct answers so far. Reading all of them I decided also to answer, since the question is known long ago and is related to a very interesting and beautifully-looking algebraic 6th degree curve - the astroid.

If a line segment with length L slides so that one end is on a x-axis, the other - on y-axis, it is well-known that the enveloping curve is an astroid (or a hypocycloid):

x^(2/3) + y^(2/3) = L^(2/3),

inscribed in a circle with radius L. I found the following movie, illustrating this (You can imagine watching it how the plumber carries the pipe around the corner):

http://xahlee.org/SpecialPlaneCurves_dir/Astroid_d...

Hence L should be chosen so that the point (2, 3) to lie on the astroid, or

2^(2/3) + 3^(2/3) = L^(2/3), or L = (2^(2/3) + 3^(2/3))^(3/2),

as in Awms' answer above.

/L = (a^(2/3) + b^(2/3))^(3/2) if the widths were a and b instead of 2 and 3/

Here is some more information:

http://en.wikipedia.org/wiki/Astroid

http://mathworld.wolfram.com/Astroid.html

/the paragraph "The astroid can also be formed . . " has a proof of the above statement about the envelope - similar approach leads to a more complicated equation if we follow Mathman's interpretation above - sliding rectangle instead of a segment/

It will depend somehow on the diameter of the pipe. For example, if we have a 2 meter diameter pipe, it will just fit through the 2 meter passage. If it's more than 3 meters long, it won't go around the corner. If it's 3 meters long, it will just fit the 3 meter passage when carried sideways.

In other words, in our 2D visualization of the problem, the pipe needs to be modelled as a rectangle, not a line segment.

For any line connecting a point on the 3 meter hallway wall with a point on the 2 meter hallway wall and passing through the inside corner of the turn, we may define x to be the distance of the 3 meter wall contact point down the 3 meter hallway. Let f(x) be the length of such a line. Then f(x)=sqrt(9+x^2)*(1+2/x). This function is minimized for x=18^(1/3), at which point f(x)=1/9*(9+18^(2/3))^(3/2) which is approximately 7.023482382. Thus a passable pipe of maximum length cannot be longer than this length. Conversely since this is a minimum any pipe of this length will pass through the turn.

The solution is best seen with a diagram. Since I'm not beside you, I'll have to expect you've drawn the expected one.

Keeping it in the plane, we see two similar triangles, whose legs are 2 by x and 6/x by 3. The hypotenuse of the big triangle is then the sum of the hypoenuses:

√(4 + x^2) + √( 36/x^2 + 9 ) = (1 + 3/x) * √(4 + x^2)

Now we need to solve for the minimum value. In doing so, we find that the derivative is

-3/x^2 * (4+x^2) / √(4+x^2) + (1 + 3/x) * x / √(4+x^2)

= 1/(x^2 * √(4+x^2) ) * (x^3 - 12)

Setting it equal to zero, we find the root to be

x = 12^(1/3)

Substituting this into (1 + 3/x) * √(4 + x^2), we get approximately

7.023482379 m.

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A slightly different method uses trigonometry, in which we want to minimize

2/sin x + 3/cos x.

In which was again differentiate, this time getting

2cos x / (sin^2 x) - 3sin x / (cos^2 x)

. = (2cos^3 x - 3sin^3 x) / (sin^2 x * cos^2 x)

Setting it equal to zero, we get

tan x = (2/3)^(1/3)

so x = arctan((2/3)^(1/3))

and then we find the hypotenuse is (once again)

7.023482379 m.

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In my mind, the top method is better, but it's up to you. By the way, both answers can be simplified to show the exact answer in a symmetric form:

( 2^(2/3) + 3^(2/3) ) ^ (3/2)

Doing this by calculus is straightfoward and but does involve solving a quartic. The way to avoid both (no calculus, no quartic) is to find that pipe length where a slight change in angle of the pipe in the corner makes virtually no difference in touching both walls. Let θ be the angle the pipe makes with the wall of the 2 meter wide hall. Then the length of the pipe is:

3/Cos(θ) + 2/Sin(θ)

Consider a tiny angle ξ between two positions of the pipe where there is virtually no change in touching both walls. Then in order that the difference at one wall is the same as the difference at the other, we must have:

(3ξ /Cos(θ))Tan(θ) = (2ξ /Sin(θ)) / Tan(θ)

This quickly gets us Tan(θ) = (2/3)^(1/'3). We then can plug this result into the first formula for the pipe length, and simplfiy:

3(1 + (2/3)^(2/3))^(3/2)

or roughly 7.02348. This agrees with the results gained by doing this "by calculus" and solving a quartic.

The length of the longest pipe is 21.07 feet. If you let x = distance from inside wall to where the pipe touches the perpendicular outside wall, then the length of the line from the perpendicular outside walls is, with help of triangle ratios: √((x+6)²+((9(x+6)/x))²) Differentiating this, setting it to 0, and solving for x, which is then plugged back into this expression, which gets you 21.07, as the minima.

Off the top of my head, about 6.4 meters.

Make a diagram. The corner of a passage is a right triangle. Find the legs and then find the hypotenuse.

The pipe can be up to, but not including the length of the hypotenuse.

We want to find the minimum of the length of all lines that pass through the interior corner point and that touch the exterior walls.

We are thus looking at minimum of tan(x) + tan (90-x). The minimum is found at 45 degrees, which gives us two isosceles triangles 3x3, 2x2

The sum of the hypotenuse of each triangle:

Sqrt(18) + sqrt(8) = 3 sqrt(2) + 4sqrt(2) = 7 Sqrt(2)

x/3 = 2/y = (2+x)/(3+y)

Thus xy = 6 and (3+y) = 3(2+x)/x

L² = (2+x)^2 + (3+y)^2

d(L²)/dx = 2(2+x) + 2(3+y)(-6/x^2)

= 2(2+x) + 2*3(2+x)(-6/x^3)

= 0 when x³ = 18

L = 7.023 m

you can hold it up vertically and so it can be any height - you just need to be able to hold the weight and it only works supposing there is no ceiling.