While searching for gift-wrapping ideas on the internet, I came across the following question on Numerade:
What is the least amount of ribbon needed to wrap around a $2''\times10''\times20''$ box as shown in the diagram?
The "expert-verified" answer was... abysmal at best so I decided to solve it myself.
It turned out to be challenging than I had expected; I flattened out the box like so:
By the triangle inequality, the ribbon must be straight lines. Then the total length can be expressed as: $$\sqrt{a^2+b^2} + \sqrt{x^2+y^2} + \sqrt{(22-x)^2 + (12-a)^2} + \sqrt{(22-b)^2 + (12-y)^2}$$
Wolfram Alpha is unable to minimize the expression. However, if I assume that the strands are placed symmetrically, i.e. $x=b$ and $y=a$, we are left with
$$2\left(\sqrt{x^2+y^2}+\sqrt{(22-x)^2+(12-y)^2}\right)$$
which has a minimum value of $4\sqrt{157}$.
Is it possible to minimize the general expression? Or is symmetry the right way to go?
I called this problem "slightly ambiguous" since the example doesn't rigorously define how to wrap, so I am open to other interpretations as well.
In the comment section, it was pointed out that $a=y$ and $b=x$ do indeed minimize the expression, so now I am looking for a synthetic proof for that fact.
EDIT (ORIGINAL ANSWER AT THE END).
Here's a quicker and neater way to get the result. I represented below the complete path of the ribbon, wrapped twice around top face (blue) and bottom face (red), once on lateral faces (orange).
We get minimal length if the path is a straight line, as in figure below, with the constraint that those two points labeled $A$ represent the same point, i.e. those two segments labeled $a$ must have the same length.
The ribbon is then the hypothenuse of a right triangle with legs $44$ and $24$ and its length is thus $4\sqrt{157}$, irrespective of the value of $a$.
ORIGINAL ANSWER.
Complete your drawing with four copies of the top rectangle and extend on them the lines representing the ribbon. The two triangles formed up-right must be the same, which implies the lines to be parallel. But also the triangles formed bottom-left must be equal, and that can happen only if $a=y$ and $b=x$.
In fact, by triangle similarity we have $b:x=a:y$, that is: $$ a=ky,\quad b=kx $$ for some number $k$. But we have also: $$ (12+a):y=(22+x):x \ \implies \ {12\over y}+k = {22\over x}+1 $$ and $$ (22+b):x=(12+y):y \ \implies \ {22\over x}+k = {12\over y}+1. $$ From the comparison of those equalities we immediately get $k=1$, hence $$ y=a,\quad x=b. $$ Moreover, we find that $$ {22\over x} = {12\over y}\ \implies \ y={6\over11}x. $$ Let's substitute that into your expression for the length $L$ of the ribbon: $$ \begin{align} L & = 2\left(\sqrt{x^2+\left({6\over11}x\right)^2} +\sqrt{(22-x)^2+\left(12-{6\over11}x\right)^2}\right)\\ & = 2\left(\sqrt{{157\over121}x^2} +\sqrt{(22-x)^2+\left({6\over11}(22-x)\right)^2}\right)\\ & = 2\left({\sqrt{157}\over11}x +{\sqrt{157}\over11}(22-x)\right)=4\sqrt{157},\\ \end{align} $$
which is thus the requested minimum!