Volume

This is really a two dimensional problem, as the same height multiplies everything. So all we need to deal with is a square and a circle.

We can arbitrarily set the radius of the circle as 1 (since we are only interested in ratios of sizes not the actual values). Area $A_{circle}=\pi$ .

Side of the square will be the diameter, 2, giving us the area $A_{square}=4$ .

So the ratio will be $\frac{4-\pi}{\pi}=\frac{4}{\pi}-1$

Solving for the volume of the cube

$V_{cube} = a^3$

Solving for the volume of the cylinder

$V_{cylinder} = A_{base}h = \frac{π}{4}(a^2)(a) = \frac{π}{4}a^3$

Solving for the volume of the wasted material

$V_{wasted-material} = V_{cube} - V_{cylinder} = a^3 - \frac{π}{4}a^3 = a^3(1 - \frac{π}{4})$

Solving for the ratio

$\frac{V_{wasted-material}}{V_{cylinder}} =$ $\frac{a^3(1-\frac{π}{4})}{\frac{π}{4}(a^3)} = \frac{4}{π} - 1$