POSSIBLE SOLUTION(S)

The general case suggests that we make a square cut of length

x = a in each corner but that we extend the edge back from the perpendicular cut by a distance y = b so that we do not have an edge which is perpendicular to the base when we fold the edges, but rather a more open box in which the top cross section is larger than the bottom cross section.

If we do this we can see that we shall need the actual height of the box when we fold the edges. This height (h) is one leg of a right triangle in which the other leg is b and the hypoteneuse is a, hence by Pythagorean Theorem we have h = Sqrt[a^2 - b^2].

Thus now we can "see" that our folded up volume consists of four kinds of pieces:

(1) the inner box of dimensions (8 - 2a), (10 - 2a), and h,

(2) two prisms of length 8 - 2a and triangle cross sections with legs

b and h,

(3) two prisms of length 10 - 2a and triangle cross sections with legs b and h, and

(4) four quarters of a pyramid each with square base b by b and height h.

We add up these volumes to get the total volume of our box.

We offer a contour plot over the entire feasible region to help locate potential maxima.

Thus we go after a maximum near a = 2 and y = b. We set the partial derivatives with respect to a and b equal to zero.

Thus, ae, the square cut edge is 2.04789 cm.

And, be, the back cut is 1.08581 cm.

The box then has a base of dimensions

and a top of dimensions

And hence the maximum volume is 61.2466 cm^3 as seen below.

We now draw a picture of the figure - presuming we lay the flat sheet on the x-y plane with one vertex at the origin and the 8 cm side along the x- and the 10 cm side along the y-axis. Pi' are the points in the flat sheet which we cut into and form the bottom vertices of the box and the Qi's are the resulting top vertices.

We examine the square cut case when we cut a square piece out of each corner

We set up the volume for a square cut of a by a from each of the corners.

And maximize this volume.

Thus we should cut in an edge of length 1.47247.

In this case we do not make a back cut.

And hence the maximum volume is 52.5138 cm^3 as seen below.

A comparision between the back cut method and the square cut method shows that the former gives greater volume.

We note that we indeed get a greater volume by permitting the back cut for the volume in that case is 61.2466 cm^3.

We now draw a picture of the figure resulting from the simple case - presuming we lay the flat sheet on the x-y plane with one vertex at the origin and the 8 cm side along the x- and the 10 cm side along the y-axis. Pi' are the points in the flat sheet which we cut into and form the bottom vertices of the box and the Qi's are the resulting top vertices.

And finally we draw both figures together to compare them.