Assume we're looking at the first quadrant in the plane with the positive $x$ and $y$ axes as boundaries. Look at a point $ (a,b) $ in the interior and consider ladders passing through this point, that is, line segments with one end on the $x$-axis, the other end on the $y$-axis and passing through $ (a,b) $. The question is this: what is the minimum length of such a ladder?
The answer is $ (a^{2/3} + b^{2/3})^{3/2} $ as can be found out with some easy calculus. But I'm wondering if there is a physical interpretation for such an answer, in particular for the exponent $ 2/3 $ and its reciprocal. The answer is neat, but the ladder which minimizes this does not have symmetrical $x$ and $y$ coordinate endpoints: $ x = a + (ab^2)^{1/3} $ and $ y = b + (a^2b)^{1/3} $, which is another curious case to ask this question.
Best Answer
For a physical interpretation, the given problem is equivalent to a rigid bar with fixed lenght constrained to move keeping end points on each of the axes and also costrained to keep the contact with the point $P=(a,b)$.
In this way, indeed, we set a stationary condition for the length of the line segment passing through $P=(a,b)$ which corresponds to its minimum value.
In such situation, for the first constraint, the instantaneous center of rotation must be located at $C=(x,y)$ and, for the second constraint, the line $CP$ must be orthogonal to the rigid bar.
Indeed, according to the following sketch
we obtain geometrically the conditions
$$\frac{x-a}{y-b}=\frac{y}{x}=\frac{b}{x-a}$$
which lead to the given solution.
Remarks
Animation for the physical analogy