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A boy in order to tidy his room asks his parents for a cardboard box to store lots of small toys. Unfortunately they didn't find any but only a raw cardboard sheet of dimensions 60cm x 80cm. Being very busy they told him to make one by himself.

What are dimensions of the biggest cuboid box volume the boy can make from given cardboard sheet?

Let assume that all adjacent faces which are not attached together naturally, should be glued together by strips of width at least 1cm

sample grid

(There is no requirement that the strip must be attached to "source" face but you will need then another 1cm to glue it to the "source" face)

separate strips

Clarification: all adjacent faces should be eventually joined together somehow (including the lid)

There is no particular requirement for the shape of joining strips. Let assume some fair minimum: a trapezoid of one side: face side length, the opposite side at least half of it and height 1cm

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  • $\begingroup$ Is there a requirement on the angles on the edges of the joining strips? It seems like it might be a possibility to make them diagonal, but that might require particular angles. $\endgroup$
    – Vicky
    Commented Mar 18, 2019 at 12:50
  • $\begingroup$ Do we assume that the joining tabs have 45 degree angles at the corners? Or are other angles allowed (like very acute angles). EDIT - I see that @Vicky asked the same thing. $\endgroup$
    – chasly - supports Monica
    Commented Mar 18, 2019 at 12:51
  • $\begingroup$ @chaslyfromUK sorry, I deleted my comment because I misunderstood your question. Please see clarification. $\endgroup$
    – mpasko256
    Commented Mar 18, 2019 at 12:54

4 Answers 4

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New solution, with larger volume:

The volume is $21000 \space cm^3$
Dimensions $x=30, \space y=25, \space z=28$

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:
$x + z = 60 - 2 = 58$
$2y + z = 80 - 2 = 78$
$2y + x = 80$

enter image description here

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  • $\begingroup$ @mpasko256 following your new problem I have found a better answer to this one. $\endgroup$
    – Weather Vane
    Commented Mar 19, 2019 at 20:42
  • $\begingroup$ Wow, I have to admit that it exceeded my expectations even with such a simple solution. Congratulation! $\endgroup$
    – mpasko256
    Commented Mar 20, 2019 at 10:42
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My solution (but see my new, second answer):

The box is 31 x 27 x 24.5 = 20506.5 cm^3 (see edit)

The cardboard cuts out to 4 pieces, the dashed lines are folds.
The two large pieces each fold to a form a U shape.
One of them has six 1 cm tabs, the other has no tabs.
There are two separate tabs, each 2cm wide.
These two tabs are slightly longer than needed, can be cut from 31 to 27.
So the only wastage is a small area $4 \times 4 = 16 \space cm^2$

enter image description here

EDIT: thanks to @mlk.

Reattach the fixed tab at the centre: from the face on its right to the face on its left.
Add a fixed tab to the bottom face on the left.
Remove one of the two loose tabs.
The face sizes become slightly larger and more equal to each other, giving a larger volume.

The box is now $30.5 \times 27.5 \times 24.75 = 20759.0625 \space cm^3$

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  • 1
    $\begingroup$ You can do with one less separate tab: If you add an attached tab at the bottom of the left piece, you can remove the attached tab on the left side of the middle part of the right piece and then use the freed space for an attached tab at the middle part of the left piece. If I haven't miscalulated, then you can change the numbers to 27.5cm, 30.5cm and 24.75cm for a volume of 20759.0625cm³ $\endgroup$
    – mlk
    Commented Mar 18, 2019 at 17:20
  • $\begingroup$ The top and bottom of the left trio attach to the 31 length tabs on the edges of the central face of the right trio, so no tab is needed where you suggest. The two loose tabs only need to be 27 long, to join edges of that length. $\endgroup$
    – Weather Vane
    Commented Mar 18, 2019 at 17:24
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    $\begingroup$ Yes, I got that, my argument was that attaching the bottom of the left trio with a tab on said bottom allows you to remove the corresponding tab on the middle face of the right trio, which in turn makes space for a tab attached to the middle face of the left trio, which saves you one of those separated tabs. However I suspect, I have to draw a diagram, communicating this in words is hard. $\endgroup$
    – mlk
    Commented Mar 18, 2019 at 17:30
  • $\begingroup$ @mlk, ah yes, I had an earlier solution, but with less volume because the largest and smallest dimensions had a greater difference. I evened up the dimensions a bit by using some horizontal space for fixed tabs. Ideally the volume is maximised when all three dimensions are equal. $\endgroup$
    – Weather Vane
    Commented Mar 18, 2019 at 17:35
  • $\begingroup$ @mlk you are absolutely right, I have edited the answer, but I don't want to redraw the image ;( $\endgroup$
    – Weather Vane
    Commented Mar 18, 2019 at 18:00
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A slight improvement on Weather Vane's solution:

solution The pieces are not to scale and actually should touch, I just drew them a bit apart to show the details. The pieces fit onto the sheet if $a+b \leq 58$, $2c+a \leq 80$ and $2c+b \leq 77$, so optimizing the volume $abc$ yields a maximum of 20759.0625cm³ for $a=30.5, b=27.5, c=24.75$.

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Disclaimer: This is not the optimized answer but it shows how to solve it:

Since we need some strips, In order to maximize it, you need to have

6 squares taken out from the 60x80 box shown

as below;

enter image description here

with the dimension

$80/3$

So we create our strips out of the rest of the pieces (6.66x80) (enough to make (strip dimensions will be (fourteen of 80/3 x 1))as a result our box volume becomes ;

$(80/3)^3=18,963$.

This could be improved a bit more by increasing one edge of the 4 squares dimensions into 4 rectangles but I leave that to someone else for now :)

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  • $\begingroup$ Yes! You are certainly going in the correct way! Congratulations! :) $\endgroup$
    – mpasko256
    Commented Mar 18, 2019 at 13:23

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