# Trash Bin Factory

Trash bins are to be fabricated from $4ft.$ x $8ft.$ Ga.#16 Aluminum sheets. To produce $1$ bin, it will take some cutting, bending and welding of a single Aluminum sheet. To maximize the capacity from the materials, the design is simply an "open box". For easier handling, the number of cut shapes must not be more than $5$.

What shall be the dimensions of the trash bin?

Basically, the question asks for the optimal way to cut a $4 \times 8$ rectangle into no more than $5$ pieces, so that you can rearrange the pieces and make an open box with maximal volume.

• No need to compute for the thickness. Just making sure the sheet is bendable and weldable. – TSLF Oct 16 '16 at 18:56
• @TSLF I have one solution but there may be more – Beastly Gerbil Oct 16 '16 at 18:59
• With the practical twist of limiting the number of pieces, this is a surprisingly interesting puzzle. Wonder why people are offering answers but not voting approval for the question. – humn Oct 16 '16 at 23:09
• The clause "One to 5 pieces of cut shape will do for easier handling." means that you can cut up to 5 times? – Matsmath Oct 17 '16 at 8:12
• @TSLF, please tip me next time before/right after posting your puzzle, so that can improve the wording and not get thumb downed so much. – Puzzle Prime Oct 17 '16 at 11:13

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:

1. This point is actually a local maxima, not a minima or other stationary point
2. A higher maxima is not attained on any other other boundary
3. 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.

• This is ingenious. Although, I am almost certain (but my memory could be wrong) that you need to address whether your extrema found by Lagrange multipliers is indeed a maximum (and not a minimum). Also, it might be a saddle point, corresponding to neither. You also do calculations on an open domain (as a,b,c>0), and I am not entirely certain what happens near the boundary. – Matsmath Oct 17 '16 at 9:21
• You should include the volume 17.4 – paparazzo Oct 17 '16 at 17:21
• I learned more from the final dissection steps here than I even understand yet – humn Oct 25 '16 at 8:47

$2ft.$ width, $2ft.$ height and $4ft.$ length

Cut and bent like this:

And

...

The two sides bend upright to form two ends, the two cut pieces form the sides

This is a total of

16 cubic feet storage space

• 4 x 2 x 2 =16 cu.ft. is good,and you can make 2 cutting pieces with that . But a little short for the maximum. – TSLF Oct 16 '16 at 19:05
• @TSLF the maximum? The question just says what are the dimensions? – Beastly Gerbil Oct 16 '16 at 19:06
• There are other ways to make the bin with larger dimensions and capacity. But not something like there are so many small parts..for practical purposes. – TSLF Oct 16 '16 at 19:19

This puzzle is easily solved with some algebra (and a bit of calculus):

- The capacity of the bin is $x^2*h$, where x is the length of a side on the bottom of the box and h is the height.
- The surface area of this open box is $x^2 + 4hx$ from the bottom of the box and 4 other sides.

I think the goal is to find the best ratio of surface area to capacity, so we want $c\over{sa}$ to be the largest possible value.
$$sa = 32S = x^2 + 4hx$$
$$\rightarrow 32S + 4h^2 = x^2 + 4hx + 4h^2$$
$$\rightarrow 2\sqrt{8S + h^2} = 2h+x$$
$$\rightarrow 2(\sqrt{8S + h^2}-h) = x$$
$${c\over{sa}} = {x^2h\over{x^2 + 4xh}} = {xh\over{x+4h}} = {2h(\sqrt{8S+4xh}-h)\over{2(\sqrt{8S+4xh}-h)+4h}} = h{(\sqrt{8S+4xh}-h)\over{\sqrt{8S+4xh}+h}}$$
Here's where I went to wolfram to get the values of $S$ and $h$ with the best ratio (I don't remember enough calculus to derive it). When $S = 5$ I got the best ratio, with $h={2\sqrt{10/3}}$. From that we get $x= 4\sqrt{10/3}$. The final dimensions in decimal are $7.3$x$7.3$x$3.65$

• ^vote with a note: Nice to know the mathematical maximum but cubical isn't the optimal shape this time. A lower wider box would take advantage of the top not counting toward surface area. – humn Oct 16 '16 at 19:15
• @waylon531-the theoretical maximum is around 3.27'x3.27'x1.63' – TSLF Oct 16 '16 at 19:27
• @humn Yeah, I assumed too much. I'm working on getting the best shape now. – waylon531 Oct 16 '16 at 19:31
• 7.3 x 7.3 from s 4 x 8 ? – paparazzo Oct 20 '16 at 20:56

You might as well considered other cut shape option as ff: 