Making a cuboid

Following Jeremy's dissection of the unit square, suppose instead the two smallest faces have dimensions $a \times b$ , while the other two pairs have dimensions $\frac{1}{2} \times a$ and $\frac{1}{2} \times b$ , so that $a+b+a=1$ where $a<b$ . The volume of the cuboid is then

$V = \dfrac{1}{2}a(1-2a)$

so that the maximum occurs at $a=\dfrac{1}{4}$ . $\;$ Any departure from this (resulting in leftover paper) will mean less volume.

Cut the paper into four equal squares. Two of them are two of the faces of size 1/2 x 1/2. Cut the other two squares in half to make four rectangles of size 1/2 x 1/4, these are the other four faces. The total volume is 1/2 x 1/2 x 1/4 = 1/16 so the answer is $\boxed{16}$ . No paper is wasted.

Note: I have not proven this is optimal, but I'm fairly confident it is.