Let's call the side lengths of the box $a$, $b$, and $c$, with $a$ being the height. Then the surface area of the box is $2ab+2ac+bc$ (four sides plus bottom). Without considering the constraint on the number of pieces, the problem becomes maximizing the volume, $V=abc$, subject to the constraints:
$$
2ab+2ac+bc\le 4\times 8=32 \\
a>0\quad b>0\quad c>0
$$
We can find the maximum using the method of Lagrange multipliers by finding the maxima of a modified "volume":
$$
\tilde V(a,b,c,\lambda)=a b c-\lambda(2ab+2ac+bc-32)
$$
$$
\begin{align}
0=\frac{\partial\tilde V}{\partial a} &= bc-2\lambda b-2\lambda c \\
0=\frac{\partial\tilde V}{\partial b} &= ac-2\lambda a-\lambda c \\
0=\frac{\partial\tilde V}{\partial c} &= ab-2\lambda a-\lambda b \\
0=\frac{\partial\tilde V}{\partial \lambda} &= 2ab+2ac+bc-32 \\
\end{align}
$$
(Note that the last equation is just our original constraint.) Factoring the equations by completing the square gives us:
$$
(b-2\lambda)(c-2\lambda)=4\lambda^2 \\
(a-\lambda)(b-2\lambda)=2\lambda^2 \\
(a-\lambda)(c-2\lambda)=2\lambda^2 \\
$$
From the last two we can see $b=c$ (so the box's bottom is square). Now we just have:
$$
(b-2\lambda)^2=(2\lambda)^2 \\
(a-\lambda)(b-2\lambda)=\lambda(2\lambda) \\
4ab+b^2=32
$$
From the first line we see $b-2\lambda=2\lambda$, so $b=4\lambda$. A similar process in the next line yields $a=2\lambda$; therefore $2a=b=c$ (so the box is half as high as it is wide). Finally, we get:
$$
4a(2a)+(2a)^2=32 \\
12a^2=32 \\
a=\sqrt{\tfrac{8}{3}}\approx 1.633 \\
b=c=2\sqrt{\tfrac{8}{3}}\approx 3.266 \\
abc=\tfrac{64}{3}\sqrt{\tfrac{2}{3}}\approx 17.419
$$
If you want to be careful, we need to check that:
- This point is actually a local maxima, not a minima or other stationary point
- A higher maxima is not attained on any other other boundary
- A higher maxima is not attained on the interior of the domain
For 2. we can see that $V=0$ on all the other boundaries. For 3. we can see that given any point on the interior, we can uniformly scale $a$, $b$, and $c$ until we reach the boundary, resulting in a larger volume.
Finally, for 1. we have to think about the general "shape" of the domain. We know that the volume is bounded (intuitively we can recognize that we can't have infinite volume with finite material); that the maxima is reached on a boundary; and we know that the volume is zero along all the boundaries except one, which has one stationary point. From this we can conclude that this stationary point must be the global maxima.
To get some intuition about the shape of the problem, here is a visual depiction of the volume on the constraint surface, from which we can see that there is indeed a single maxima:
The dimensions of the bottom of the box are $2a\times 2a$, and the sides are $1a\times 2a$. This gives us a total area of $12a^2$, which we can rearrange into a $3a\times 4a$ rectangle.
Next, we need to find a dissection of this rectangle into a $4\times 8$ rectangle; one way is shown below:
Based on the above dissection, my solution is as follows:
We start with the $4\times 8$ sheet on the left. We cut it into four pieces along the solid lines and reassemble them into the shape on the right using three welds. Finally, we bend the shape upwards along the dashed lines and make four more welds to form the shape into a box.
