Fitting a dodecahedron inside a cube I'm afraid I'm not fantastic at maths and am struggling with a problem.
I am a woodworker and have been asked to cut a solid dodecahedron from a 3 inch cube of wood. I am struggling to figure out what the maximum sized dodecahedron I could make is.
Is there a formula for calculating this?
 A: The most "natural" dodecahedron positioning is by placing two of its vertices on each of the $6$ faces of the cube, as follows:

(In the above image, I've bolded the edges which are tangent to the faces of the cube. Each of them passes through the center of their face.)
This is probably the way that I would try to cut a dodecahedron out of a cube, because every face in this dodecahedron is at a right angle to a face of the cube. This means that you can make easy vertical cuts to produce these faces by setting the cube flat on one of its faces.
However, it's not the largest dodecahedron possible:
                                          
This image, from Moritz Firsching's 2018 paper Computing maximal copies of polytopes contained in a polytope (PDF), depicts a maximal dodecahedron inside a cube. It has a side length of around $0.3943$ times that of the cube, compared to the original diagram's $\phi^{-2}\approx 0.382$.
Given the marginal improvement for what I assume is a much more difficult woodworking experience, I wouldn't particularly recommend it.
For an approach starting with a more cylindrical piece of wood (namely, a tree stump), see this IRL guide.
