# Make the largest box from a cardboard sheet Chapter #2

Please see: Make the largest box from a cardboard sheet

Thanks to his older brother's friends: @Oray, @Weather Vane and @mlk, the boy managed to make as large cardboard box as possible. Unfortunately, it was not enough to store all his toys responsible for causing mess in his room.

To do something about it, he asked his parents for another sheet to make another box. Unfortunately his father just cut off a (little bit larger) piece from remaining sheet already and used it to temporarily patch a hole in his greenhouse, so only 60cm x 39cm piece still remains.

Now the problem is similar, what is the largest box the boy can make using 60cm x 39cm cardboard piece?

Rules are the same as previously.

Edit: My third solution:

The volume is $$6944.4375 \space cm^3$$

The cardboard cuts into 2 pieces.
One folds forwards, the other folds backwards.

The box dimensions are from:
$$x + z = 39 - 1 = 38$$
$$2z = 39$$
$$2y + x = 60 - 3 = 57$$

Which gives $$x=18.5, \space y=19.25, \space z=19.5$$

Not to scale

My second solution:

The volume is $$6930.5625 \space cm^3$$

The cardboard cuts into 2 pieces.
Two of the box dimensions are the same.

The dimensions are from:
$$2A = 39 - 2 = 37$$
$$2B + A = 60 - 1 = 59$$

Giving $$A=18.5, \space B=20.25$$

Not to scale

The theoretical maximum volume will be a cube:

There can be at most 4 folds between sides, so there must be 8 tabs.
The total area = $$39 \times\ 60 = 2340$$
Which should be made up from 6 faces and 8 tabs.

Suppose the side length is $$s$$
Then $$6 s^2 + 8 s = 2340$$
Which solves as $$s = 19.093$$
Maximum possible volume = $$6960.213$$

• Good job, but I still see some room for improvements (~10cm^3). Your answer is going now definitely in a good direction! ;) – mpasko256 Mar 20 at 15:25
• @mpasko256 I've found another $13.875 \space cm^3$ volume. – Weather Vane Mar 20 at 20:28
• This is the correct answer, congratulations! ;) – mpasko256 Mar 21 at 9:30
• It was satisfying to find a symmetrical solution and it's not far short of the theoretical maximum. Apart from the mitres, only four 1 cm squares are wasted. – Weather Vane Mar 21 at 9:34

My solution (a better one was since found):

The volume is $$6910.3125 \space cm^3$$
Dimensions $$x=19.5, \space y=20.25, \space z=17.5$$

The cardboard cuts into 2 pieces.
Each piece folds to a form a U shape.
One of them has eight $$1 \space cm$$ tabs, the other has none.

The box dimensions are from:
$$x + z = 39 - 2 = 37$$
$$2y + z = 60 - 2 = 58$$
$$2y + x = 60$$

• Nice try, but unfortunately this time your pattern from previous chapter does not lead to optimal solution. – mpasko256 Mar 20 at 10:27
• Hm... but it did lead back to a better solution for puzzle #1 where my new answer has a larger box than the accepted answer. – Weather Vane Mar 20 at 10:30
• Yes but please notice difference in proportions. If we compare everything to 2x3, previously our rectangle was shorter (6x8 -> 3x4), now it is narrower (a little bit less than 2 versus 3). – mpasko256 Mar 20 at 10:35
• Yes, I realise the optimal solution will be the closest to x = y = z but with only four 1 cm squares wastage here (apart from the mitres), and no loose tabs, I thought it would be a good shot. – Weather Vane Mar 20 at 10:37

Adapting mlk's accepted answer to the previous question, using the same basic template, we now need:

$$a+b \leq 37$$, $$2c+a \leq 60$$ and $$2c+b \leq 57$$

So one possible template would be identical to the one in that answer but using

a = 20, b = 17, c = 20, for a total volume of 6800.

As this is essentially the same as the known answer to the previous question, I assume it represents a baseline from which a better answer tailored to this question will improve.

One way to improve the volume would be

Try to make a perfect cube. a = b = c = 19, for a total volume of 6859 seems plausible, but the same template layout as the previous answer no longer works...

I initially thought a layout using this idea was possible, but didn't manage to put anything together before Weather Vane provided a far better solution, which seems likely to be optimal, so I'll stick with the baseline above as my non-optimal borrowed answer.