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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

(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)

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

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

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 $4 times 4 = 16 space cm^2$

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 $30.5 times 27.5 times 24.75 = 20759.0625 space cm^3$

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;

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 :)

A slight improvement on Weather Vane's 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 $a=30.5, b=27.5, c=24.75$.