Intuitive explanation for formula of maximum length of a pipe moving around a corner?

Question

For one of my homework problems, we had to try and find the maximum possible length $L$ of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths $A$ and $B$ (assuming everything is 2d, not 3d):

My professor walked us through how to derive a formula for the maximum possible length of the pipe, ultimately arriving at the equation $L = (A^{2/3} + B^{2/3})^{3/2}$.

The issue I have is understanding intuitively why this formula works, and exactly what it's doing. I understand the steps taken to get to this point, but there's an odd symmetry to the end result -- for example, is the fact that $\frac{2}{3}$ and its inverse are the only constants used just a coincidence, or indicative of some deeper relationship?

I also don't quite understand how the formula relates, geometrically, to the diagram. If I hadn't traced the steps myself, I would have never guessed that the formula was in any way related to the original problem.

If possible, can somebody give an intuitive explanation as to why this formula works, and how to interpret it geometrically?

---

Here's how he found the formula, if it's useful:

The formula is found by finding the maximum possible length of the pipe by expressing the length in terms of the angle $\theta$ formed between the pipe and the wall, and by taking the derivative to find when $\frac{dL}{d\theta} = 0$, which is the minimum of $\frac{dL}{d\theta}$ and is therefore when $L$ is the smallest:

$$ L = \min_{0 \leq \theta \leq \frac{\pi}{2}} \frac{A}{\cos{\theta}} + \frac{B}{\sin{\theta}} \\ 0 = \frac{dL}{d\theta} = \frac{A\sin{\theta}}{\cos^2{\theta}} - \frac{B\cos{\theta}}{\sin^2{\theta}} \\ 0 = \frac{A\sin^3{\theta} - B\cos^3{\theta}}{\sin^2{\theta}\cos^2{\theta}} \\ 0 = A\sin^3{\theta} - B\cos^3{\theta} \\ \frac{B}{A} = \tan^3{\theta} \\ \theta = \arctan{\left( \frac{B}{A} \right)^{\frac{1}{3}}} \\ $$

At this point, we can substitute $\theta$ back into the original equation for $L$ by interpreting $A^{1/3}$ and $B^{1/3}$ as sides of a triangle with angle $\theta$ and hypotenuse $\sqrt{A^{2/3} + B^{2/3} }$:

$$ \cos{\theta} = \frac{A^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \sin{\theta} = \frac{B^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \therefore L = A^{2/3} \sqrt{A^{2/3} + B^{2/3} } + B^{2/3} \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3}) \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3})^{3/2} \\ $$

The equation for the formula for the maximum length of the pipe is therefore $L = (A^{2/3} + B^{2/3})^{3/2}$.

Answer 1

If this result could be explained intuitively it wouldn't have been necessary to go through the calculations. A priori you could expect $L$ to be homogeneous of degree $1$ in $A$ and $B$, as well as being symmetric in $A$ and $B$. As a matter of fact the result of the calculation could as well have been $$L=2\min\{A,B\},\quad{\rm or}\quad L=\sqrt{A^2+AB+B^2}\ ,$$ or an expression involving some other exponent in place of ${2\over3}$.

Answer 2

As far as I know, this isn't a case where there's an intuitive explanation of this problem on its own. Here are two substitutes:

First, there are simpler calculations using more advanced techniques, and maybe these calculations will make the form of the answer seem less mysterious. For example, if we rephrase the question as

Find the length of the shortest line segment from $(x,0)$ to $(0,y)$ that passes through $(a,b)$.

then with a little coordinate geometry we can rephrase it again as

Minimize $(x^2+y^2)^{1/2}$ subject to the constraint $\frac ax + \frac by = 1$.

Using the method of Lagrange multipliers from multivariable calculus on this version of the problem yields quickly that the optimum has $(x^3,y^3)$ proportional to $(a,b)$.

Second, there is a bigger picture about this kind of problem — if we rephrase the question again by replacing $x,y$ with their reciprocals, we get the $p=-2$ case of the problem

Minimize $(x^p+y^p)^{-1/p}$ subject to the constraint $ax+by=1$.

Expressions like $(x^p+y^p)^{1/p}$ can be thought of as a more general notion of "distance". With $p=2$ it's the usual distance, as in Pythagoras' theorem; with $p=1$ you get taxicab distance; for $p\ge 1$ you get various weird notions of "distance" which still act in some essential ways like the familiar notion; for $0<p<1$ you get notions which are still somewhat distance-ish, but not very; for $p<0$ the feeling of "distance" has been left far behind, but we continue with stout hearts. (More.)

The relationship between the $p=-2$ notion of "distance" which occurs in the (rephrased) problem statement and the $p=\frac23$ notion of "distance" which occurs in the solution is a much studied one; these notions of distance are said to be "dual" to each other. For people who have studied this subject, the outcome here is quite natural — it's "because" $\frac23$ is the conjugate exponent of $-2$. (We need a version of Hölder's inequality for negative exponents, which is not always taught, but it exists.)

(The notion of "duality" between notions of distance has its roots in 19th century geometry, in efforts to understand geometric shapes not just as collections of points but also as envelopes of their tangents. From this point of view, we could say that your problem is dual to this one: if a ladder standing straight against a wall slides down to the ground (its top sliding down the wall and its base sliding along the ground), what curve is tangent to the ladder at all times? See Astroid, where you will see your $\frac23$.)

So, no, the form of the answer is not a coincidence, but as far as I know it is only "intuitive" in retrospect, which really means only that it has become familiar.