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### Maximizing the Volume of a Box

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Date: 11/05/96 at 19:41:49
From: Anonymous
Subject: Rectangular box

I need the formula or equation which will solve the following problem:

I have a two-dimensional rectangular piece of paper 20 by 10 and I
want to make it into a box with the greatest possible volume. I cut
out a square in each corner and fold up the sides. The size of my
square in this case is 2.11 by 2.11 (worked out by trial and error).
If my rectangle is 30 by 10 the square is 2.26 by 2.26 (also by trial
and error).  There must be a formula but I cannot find it.

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Date: 01/11/97 at 15:42:14
From: Doctor Donald
Subject: Re: Rectangular box

In fact, this is a pretty tough problem. It is good that you have
worked out some trial-and-error results. If you were just interested
in solving the problem for a sheet of paper of one size, then you
that describes the volume of the box (which may have been how you got

Using calculus, we can get a formula which, when you put in the
dimensions of your paper, will give you the size of the squares that
you should cut out.

Suppose your paper is L by W, with L >= W, and you cut a square of
size x out of each corner.  Then as you probably already know, the
volume of the box you get is:

V(x) = x(L-2x)(W-2x)

since the box is x deep, L-2x long, and W-2x wide.

Multiply out to get V(x) = xLW - 2x^2(L+W) + 4x^3

The derivative of V is

V'(x) = LW - 4x(L+W) + 12x^2,

and this has to be 0 when V has a maximum.  The domain of V is
0<=x<=W/2, since none of the dimensions of the box can be negative,
and any value of x > W/2 will make the box width negative.

Set V'=0 and solve using the quadratic formula:

x = (4(L+W) +- sqrt((4(L+W)^2 - 48LW))/(24

There are two solutions here, and one of them is outside the domain.
We need to do some simplification so we can see what's up.  Multiply
out inside the radical to get:

x = (4(L+W) +- sqrt(16L^2 + 16W^2 - 16LW))/24

Remove the common factor of 4, cancel with the 24, and get:

x = (L + W +- sqrt(L^2 + W^2 - LW))/6

Now which root do we want?  It's not easy to see what to look at, but
remember L>=W.  The thing under the radical is at least as big as W^2,
since L^2 - LW = L(L-W) >= 0.  So if we ADD the sqrt, the numerator
is >= W + W + sqrt(W^2) = 3W, and so

x > 3W/6 = W/2.

This is no good since it is outside the domain, so we subtract
instead.  The only useful root is:

x = (L+W - sqrt(L^2 + W^2 - LW))/6

and this is the formula we want, since the only other possibility is
that V has a max for x = 0 or x = W/2, and both of these values for x
give a volume of 0 for the box.

If you try this formula for L = 20, w = 10 you get your trial and
error result of about 2.11.  If you use L = 30 and W = 10 you get

-Doctor Donald,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus

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