I'm planning on building a storage box out of sheet wood, and I'd like to know how to cut up my sheet into 6 pieces to form a box with the largest volume possible. Ideally the box should be flush on the outside, but I imagine the problem is a lot easier to solve when not taking that into account.
In the case of a non-flush box I'd like a solution that takes the width and height of the sheet and gives me the width height and depth of the interior of the resulting box.
On
The answer will depend in part on the aspect ratio of the original sheet of wood, that is, the ratio of length to width.
For two sheets of the same aspect ratio but different sizes, that is, two similar rectangles, you can take any plan for cutting one sheet and assembling a box, and scale it up or down to build a similar box with the other sheet. An optimal plan for a box using one sheet translates in this way into an optimal plan for the other sheet.
If one sheet is $x$ times as wide as the other, its area will be $x^2$ as great as that of the other sheet, but the volume of the similar box will be $x^3$ as great as that of the other box. For any sheet of a given aspect ratio, the volume of the optimal box will be some constant times the $3/2$ power of the area of the sheet, that is, $V_{\max} = k_{\max} A^{3/2}$, where $k_{\max}$ is a best-case multiplier determined by your sheet's aspect ratio.
The question then is how to find $k_{\max}$, given an aspect ratio.
The theoretical best case
The best value of $k_{\max}$ that any sheet of wood could possibly have would occur in the cases where you are able to cut a thin sheet so as to build a cube with no wood discarded.
Assuming an extremely thin sheet, a sheet of dimensions either $w \times 6w$ or $2w \times 3w$ can be cut into six squares of side $w$, which then enclose a cube of side $w$ and volume $V = w^3$. The original area of the sheet was $A = 6w^2$, giving us the relation $\sqrt[3]{\strut V} = \sqrt{\frac16 A}$, so $$ V = \frac{\sqrt{6}}{36} A^{3/2} $$ Since a cube is the rectangular box of maximum value for a given surface area, and since these plans use all the area of the original sheet, the factor $\sqrt 6/36 \approx 0.06804138174$ is the best possible such factor for any sheet of any aspect ratio.
But you can do this only with the two aspect ratios $1:6$ or $2:3$. You cannot build a cube without waste from any other sheet of wood, so for any other sheet you will be forced to use a different pattern and if you are able to determine the value of $k_{\max}$, you will find that $k_{\max} < \sqrt 6/36$.
A more practical example
Sheets of wood seem more likely to come with an aspect ratio of $1:2$ rather than $1:6$ or $2:3$; a typical sheet for construction in seems to be $4$ feet wide by $8$ feet long. (Even in countries using the metric system, it seems the size may be given in metric but will be $4\ \text{feet}\times8\ \text{feet}$ to within less than $\frac1{10}\%$ in each dimension; for our purposes, however, only the aspect ratio matters.) But rather than $4$ feet or $1.2$ meters, let's just say the width of the sheet of wood is $w$ and its length is $2w$.
A likely plan for a sheet of wood of dimensions $w \times 2w$ (area $A = 2w^2$) is to cut it lengthwise in half, and then cut three pieces of dimensions $h \times \frac12 w$, $h \times \frac12 w$, and $2(w - h) \times \frac12 w$ from each of the two halves. If $h = \frac34 w$, for example, the resulting box would have dimensions $\frac34 w \times \frac12 w \times \frac12 w$ and volume $$ V = \frac{3}{16} w^3 = \frac{3\sqrt 2}{64} A^{3/2} \approx 0.06629126073 A^{3/2}. $$ This is almost as good as the theoretical best $k_{max}$ for sheets of any aspect ratio. But can we do better? What about other values of $h$?
If we let $h$ be less than $\frac34 w$, after cutting off two pieces of width $h$ from each half of the sheet, the dimension $2(w-h)$ of the remaining pieces would be greater than $\frac12 w$. In order to use the four $h \times \frac12 w$ pieces as "sides" of the box, we would have to cut down the top and bottom to $\frac12 w \times \frac12 w$, and the resulting box would have volume $V = \frac14 w^2h < \frac{3}{16} w^3$, less than when $h = \frac34 w$. So we can reject any $h < \frac34 w$.
If we let $h$ be greater than $\frac34 w$, the width of the "top" of the box, $2(w-h)$, will be less than $\frac12 w$. We can still build a box of height $h$ using the $h \times \frac12 w$ pieces as "sides", but two of the "side" pieces will be incompletely used. The box will have dimensions $h \times 2(w-h) \times \frac12 w$. As a function of $h$, the volume will be $V = (w - h)wh = w^2 h - w h^2.$ Over the interval $\frac34 w \leq h \leq w $, this function reaches its maximum value at $h = \frac34 w$.
That is, for any box made according to this pattern, the volume of the box is maximized at $h = \frac34 w$, with volume $V = \frac{3\sqrt 2}{64} A^{3/2}.$
Here's another attempt to build a better box from the $w \times 2w$ sheet:
An alternative plan for building a box from the $w \times 2w$ sheet is to cut it widthwise into pieces of dimensions $y \times w$, $(w - \frac12 y) \times w$, and $(w - \frac12 y) \times w$, where $\frac12 w \leq y < w$. We then cut two $y \times (w - y)$ pieces from the $y \times w$ piece, discarding the part left over, and cut a $(w - \frac12 y) \times (w -y)$ piece from each of the remaining pieces, leaving a $(w - \frac12 y) \times y$ piece. The resulting box has dimensions $y \times (w - y) \times (w - \frac12 y)$, with volume $V = \frac12 y^3 - \frac32 wy^2 + w^2 y$. Over the interval $\frac12 w \leq y < w$, however, the volume of the box is maximized at $y = \frac12 w$, where it turns out that $$ V = \frac{3}{16} w^3 = \frac{3\sqrt 2}{64} A^{3/2}, $$ the same maximum volume (in fact the same pieces) as in the other plan.
For a $4\ \text{feet}\times8\ \text{feet}$ sheet of plywood this results in a box of dimensions $2\ \text{feet}\times2\ \text{feet}\times3\ \text{feet}$ with volume $12\ \text{feet}^3$ or about $0.339802$ cubic meters.
This does not prove there is no better way to build the box (that is, I am not claiming that $k_{max}$ is $3\sqrt 2/64$ for a sheet of aspect ratio $1:2$), but I have not yet found a better way, and the volume achieved is fairly close to the theoretical maximum volume of any box built from a sheet with the given area.
The general case
A more general formula would take any aspect ratio, reduce it to a single number (length divided by width), and then express $k_{max}$ as a function of that number. The tricky thing about such a function is that there is no single "pattern" for cutting the wood that will be best for sheets of all shapes. (Notice how very differently we cut a $1:6$ sheet and a $2:3$ sheet, even though they end up with the exact same value of $k_{max}$.) We have not even begun to look at all the possible aspect ratios (such as $1:3$, $4:5$, $1:\frac12(1+\sqrt5)$, and so forth); and even for the $1:2$ sheet we have not yet (at least not in any answer to this question) gotten a proven value of $k_{max}$, only a "pretty good" multiplier that is at least close to if not equal to $k_{max}$ for the aspect ratio $1:2$.
On
A 4x8 sheet is 32 sq feet. For a maximum theoretical volume you would picka cube with equal sides all L^2 in area. 6xL^2=32, or L=SQRT(32/6)=2.3 feet. The resulting volume is L^3, or 12.32 cubic feet. Of course, you can't cut the 4x8 sheet into 6 equal 2.3 ft squares, so split the sheet the long way, getting pieces as close to 2.3 ft (in this case 2 ft). This will allow two ends at 2x2, then 4 equal pieces (tops and sides at 2x3). The resulting volume is then 2x2x3, or 12 cubic feet, which is as close to the maximum you will get with a realistic cutting pattern.
On
For simplicity (and practical purposes) I will just ignore the saw kerf. Generally speaking, if you are making something out lower grade material it will come in 48 x 96" sheets. Cabinetry type sheets often come in 49 x 97 sheets to account for trim cuts of slightly damaged edges and some saw kerf from your other cuts. Either way, if you use lumber to case the edges as described below, the edges of the box don't necessarily have to meet perfectly.
Rip the 4 x 8 sheet of plywood in half to make two $2 \times 8$ pieces. From both of these strips cut two 3 ft pieces and you are left with one piece 2 x 2 from the end. This will give you 4 pieces @ 3 x 2 for the sides, top and bottom and 2 pieces at 2 x 2 for the ends.
If you want to put stuff in the box, I would probably then use some sort of lumber for all of the outside edge (if you want to put things on the box you can move the lumber to the inside of the box). The width of the lumber depends on what you want to put in the box really, 1 x 2 strapping or 1 x 4 lumber would probably work for the majority of stuff. If you have some scrap 2 x 4 lying around you can also make "feet" for the bottom.
This looks like a perfectly acceptable question. A man has a sheet of material (width W and height H). What is the maximmum volume of a closed box he can make?
We can ignore the thickness of the material initially.
A square box unfolded would be 4 units by 3 units in size - but this would produce a large amount of waste. To reduce the waste, simply divide the sheet into 6 equal sized pieces ( all W/2 x H/3 in size ).
Of the six pieces, four can remain this size but the top and bottom can only be the smallest dimension square. If we assume W < H, then the remaining 2 sides are W/2 square.
Our volume therefore would be, the base area times the height, or (W/2) * (W/2) * H/3.
If this were a 2.4x1.2 sheet of ply, the maximum volume would be: (.6*.6) * .8 = 0.48 cubic metres. The waste would be 2 strips of 0.6x0.2
While this produces some waste, we can't distribute that waste evenly to the other 4 sides ( altering one dimension on a box results in four sides changing size). In this woodworking exercise, some of the excess would be used to in the top and bottom because the external dimensions would be 0.6 x (0.6 + 2 x thickness).