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Maximizing the Volume of a Box

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

Can you help me please?

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. 

Please help!

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 
could get an answer (your answer, in fact) just by graphing a function 
that describes the volume of the box (which may have been how you got 
your answer).  

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 
2.26, your other answer.

-Doctor Donald,  The Math Forum
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Associated Topics:
High School Calculus

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