A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?
$\textbf{(A)}\ 4+4\sqrt{2} \qquad \textbf{(B)}\ 2+4\sqrt{2}+2\sqrt{3} \qquad \textbf{(C)}\ 2+3\sqrt{2}+3\sqrt{3} \qquad \textbf{(D)}\ 4\sqrt{2}+4\sqrt{3} \qquad \textbf{(E)}\ 3\sqrt{2}+5\sqrt{3}$
The distance of an interior diagonal in this cube is $\sqrt{3}$ and the distance of a diagonal on one of the square faces is $\sqrt{2}$. It is not possible for the fly to travel any interior diagonal twice, as then it would visit a corner more than once. So, the final sum can have at most  $4$ as the coefficient of $\sqrt{3}$. The other $4$ paths taken can be across a diagonal on one of the faces (try to find a such path!), so the maximum distance traveled is $\textbf{(D)}\ 4\sqrt{2}+4\sqrt{3}$.