Maximizing the Volume of a Box: Find Size of Square Cutout

Date: 07/09/99 at 14:45:13
From: Anonymous
Subject: Relation between maximum volume and length/width

We are currently investigating the length of square cutout taken from 
each corner of a square/rectangle that gives a maximum volume for 
resultant box as the sides are folded up. I have found that for each 
rectangle with side ratios of 1:1, 1:2, 1:3, etc. there is a specific 
number that you divide the width of the rectangle by to give the 
length of the cutout that maximizes the box volume.

     1:1 - 6
     1:2 - approx. 4.7
     1:3 - approx. 4.4

I tried to substitute 6 (for square) for x in the formula to give 
the volume of the box, i.e. x(l-2x)(l-2x). This rounded down to 
2/27 (length cubed).

I am now trying to find a general formula for the cutout that gives 
the maximum volume of the box for a rectangle of any width-to-length 
ratio. What do you suggest? I do not understand where the set 
denominator comes from. Do you know? Thank you for any help you can 
give.

Anonymous

Date: 07/09/99 at 18:20:12
From: Doctor Anthony
Subject: Re: Relation between maximum volume and length/width

Open Box 
--------
I have given a general solution to the problem with initial dimensions 
of the sheet a * b, and having found an expression for the volume in 
terms of x, the size of the cut, have differentiated to find the value 
of x that produces the box of maximum volume. If you are not familiar 
with calculus techniques, the value of x to give maximum volume can be 
found by plotting a graph of V against x and seeing where the volume 
is a maximum.

If the card has dimensions a times b, and if x is the side of the 
square cut from each corner before folding up the sides, then the 
volume of the box is:

     V = x(a-2x)(b-2x)

       = x(ab - 2x(a+b) + 4x^2)

       = abx - 2(a+b)x^2 + 4x^3

To find a maximum point we can either plot a graph of V against x, or 
we get the maximum point by differentiating and equating to zero.

     dV/dx = ab - 4(a+b)x + 12x^2 = 0

You can use the quadratic formula to solve this equation

           4(a+b) - sqrt(16(a+b)^2 - 48ab)
     x =  ---------------------------------
                       24

           (a+b) - sqrt((a+b)^2 - 3ab)
     x =  -----------------------------
                       6

           (a+b) - sqrt(a^2 + 2ab + b^2 - 3ab)
     x =  -------------------------------------
                         6

              (a+b) - sqrt(a^2 -ab + b^2)
        x =  -----------------------------
                           6

Even if you don't understand the above, use this formula (having put 
in the values of a and b) to see what is the best value of x. You 
could then plot the graph of V against x using a few values of x 
either side of this optimum value. You will find that the graph does 
peak at the value of x found by this formula.

If a = b (square card) then:

           2a - sqrt(a^2-a^2+a^2)     2a - sqrt(a^2)
     x  =  ----------------------  =  --------------
                    6                       6

            a       side of square
        =  ---  =  ----------------
            6             6

     V  =  x(a-2x)(a-2x)  =  (a/6)(2a/3)(2a/3)  =  2a^3/27

For rectangular cards we proceed as follows:

Suppose for example you have a card of 18 * 12 inches, then a = 18,
b = 12 and the value of x will be

            30 - sqrt(324-216+144)
     x  =  ------------------------
                      6

            30 - sqrt(252)      30 - 6.sqrt(7)
     x  =  ----------------  =  --------------
                 6                    6

     x  =  5 - sqrt(7)

        =  2.354 inches

So you would cut a square of side 2.354 from each corner and then fold 
up the edges. This would provide a box of maximum volume.

     V  =  x(a-2x)(b-2x)

        =  2.354(18-4.708)(12-4.708)

        =  2.354 * 13.292 * 7.292

        =  228.162 in^3

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/