A fun Valentine's day chocolate box optimization problem...

My intuition says that this should be the diagram.

This means that the optimal band will not cross the top at a $45^{\circ}$ angle, unless the box has a square top. Otherwise, the line has a slope equal to $\frac {w+h}{l+h}$ and has total length.

The length marked off by the right triangle goes from the middle of the front tot the middle of the side. Its length is $\sqrt { (\frac {w+h}{2})^2 + (\frac {l+h}{2})^2}$ and is $\frac 14$ the total length of the band.

$2\sqrt {(l+h)^2 + (w+h)^2}$

On further thought... doesn't change the answer, but it does introduce a small nuance.

The band travels across the top twice and the bottom twice and each side once.

Figure 2.

The shortest distance is a straight line, and the dimensions the line must cross is $(2h + 2w) \times (2h + 2l)$

However, there is nothing that says the shortest distance is in fact symmetric in the way it crosses top and bottom.