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### Open Box Problem

```Date: 06/22/2003 at 16:46:14
From: Georgia Wright
Subject: Open box problem

I am not sure how to find the formula for the greatest volume box you
can make from a sheet of cardboard with different-sized corners cut
out of it.

You need calculus to solve it but I am only 13 so I need some help.

i have figured that for a sheet of card that iis n by n that the
formula is 2times n^3 all divided by 27 i found this out simply by
finding the biggest cmsquared you need to cut out of a 20 by 20
square (which is 3.33333) and simplifying (L-2L/6)(L-2L/6)L/6
with L being the length of the sheet of card, i know there is an
easier way to solve it, also a way to solve it for a rectangular
hard to follow , and i am not sure how to apply this to my problem,
i would be very grateful if you could help me it will help me get my
```

```
Date: 06/22/2003 at 21:05:38
From: Doctor Ian
Subject: Re: Open box problem

Hi Georgia,

You don't actually _need_ calculus to solve a problem like this.  It
just makes the solution easier to find.

Suppose we have a card that is L by W, and we want to cut some square
corners (of length x by x) out so we can fold it into a box:

+-------------+        ---
|             |         |
+--+             +--+      |
|                   |      | W
+--+             +--+      |
|             | x       |
+-------------+        ---

|--------L----------|

The volume of the box will be

volume = height * area of base

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

Now, if you choose particular values for L and W (e.g., L=10 and W=6),
then you get an equation that gives the volume as a function of x:

volume = x * (10 - 2x) * (6 - 2x)

Suppose we plot volume for various values of x? What will the graph
look like?

For one thing, we know that the volume will be zero when x=0. Do you
see why? It will also be zero when x=5, and when x=3. So we can plot
those points right away:

|
-
|
-
|
-
|
-
|
-
|
--*--|--|--*--|--*--|--|--|--|--|--|--|--

Of course, x can't get all the way to 5. We can't make x larger than
half the length of the shorter side of the box. And we can't make x
negative. So we really only care about values of x between x = 0 and
x = 3:

|
-
|
-
|
-
|
-
|
-
|
--*--|--|--*

So, what happens in between? Let's try a couple of values. When x = 1,
we get

volume = 1 * (10 - 2*1) * (6 - 2*1)

= 1 * 8 * 4

= 32

And when x = 2, we get

volume = 2 * (10 - 2*2) * (6 - 2*2)

= 2 * 6 * 2

= 24

Let's plot those two points:

|
50 -
|
40 -
|  *
30 -
|     *
20 -
|
10 -
|
--*--|--|--*
1  2  3

How about right in the middle, at x = 1.5?

volume = 1.5 * (10 - 2*1.5) * (6 - 2*1.5)

= 1.5 * 7 * 3

= 31.5

|
50 -
|
40 -
|   *
30 -     *
|       *
20 -
|
10 -
|
--*---|---|---*
1   2   3

How about closer to the left side of the range, at x = 0.5?

volume = 0.5 * (10 - 2*0.5) * (6 - 2*0.5)

= 0.5 * 9 * 5

= 22.5
|
50 -
|
40 -
|   *
30 -     *
| *     *
20 -
|
10 -
|
--*---|---|---*
1   2   3

Now, if you keep plotting points, you'll find that you end up with a
curve that looks kind of like a distorted parabola. (If it were
actually a parabola, we would know from symmetry that the highest
point would be halfway between x = 0 and x = 3, at x = 1.5. Clearly
that isn't the case here.)

The point of all this is that the highest point on this curve will
occur at the value of x that gives you the greatest volume for given
values of L and W. Does that make sense? So you can solve the problem
by completing the graph, without using calculus.

Just to give you a glimpse ahead, at any point P on a curve, we could
choose two nearby points A and B (also on the curve) and compute the
slope of the line segment between them, as shown below:

A
.
.
*.                             change in y
P . change in y         slope = -----------
*  .                             change in x
B.......
change in x

For most curves that we deal with, as the points A and B get closer to
P, the slope of the segment AB approaches the slope of the curve
itself at P. And here's a very useful fact: at the very top of a curve
like our distorted parabola, this slope will be horizontal, that is,
equal to zero:

P
A.....B

*             *

*                   *

And this is where calculus comes into it. If we compute the
'derivative' of the function

v(x) = x * (10 - 2x) * (6 - 2x)

= x * (60 - 12x - 20x + 4x^2)

= x * (4x^2 - 32x + 60)

= 4x^3 - 32x^2 + 60x

we get

3-1        2-1         1-1
dv/dx = 3*4x    - 2*32x     + 1*60x

= 12x^2 - 64x + 60

This equation tells us the _slope_ of the curve at any point. If we'd
like to know the values of x where the slope is horizontal - that is,
where the curve reaches a maximum (or minimum) - we can set the
derivative equal to zero and solve for x.

0 = 12x^2 - 64x + 60

0 = 3x^2 - 16x + 15

-(-16) +/- sqrt((-16)^2 - 4(3)(15))
x = -----------------------------------
2(3)

16 +/- sqrt(256 - 180)
x = ----------------------
6

16 +/- sqrt(76)
x = ---------------
6

16 +/- 8.7
x = ----------
6

= 24.7/6 or 7.3/6

The first of these is out of range (i.e., it's greater than half the
width of the box). So the second must be the value of x that we're
looking for.

Substituting it back into our equation,

volume = 1.2 * (10 - 2*1.2) * (6 - 2*1.2)

= 1.2 * 7.6 * 3.6

= 32.8

which is higher than any of the other values we've tried.

In fact, this is the same point you'd find by completing the graph.
But as you can see, this is a lot quicker, and this illustrates one
of the most important uses of differential calculus. That is, if you
can find a way to express one quantity as a function of another, you
can use derivatives to quickly find the largest or smallest possible
values of the function.

Since we're often interested in maximizing quantities (like the amount
of money that we can charge for something) or minimizing quantities
(like the amount of money that we'll have to pay for something), these
kinds of problems are called 'optimization' problems. In the case of
the box, we're trying to find the 'optimal' value of x, i.e., the one
that gives us the greatest volume.

Note that we picked specific values for L and W. But you said you were
looking for a formula that would cover any case. Can we find one?
Without using calculus, it would be awfully hard. But using calculus,
we can do this:

1) Expand the formula to get a polynomial.

volume = x * (L - 2x) * (W - 2x)

= x * (LW - 2Wx - 2Lx + 4x^2)

= x * (4x^2 - 2(L+W)x + LW)

= 4x^3 - 2(L+W)x^2 + LWx

2) Find the derivative and set it equal to zero.

d volume
-------- = 12x^2 - 4(L+W)x + LW
dx

0 = 12x^2 - 4(L+W)x + LW

-[-4(L+W)] +/- sqrt([-4(L+W)]^2 - 4[12][LW])
x = --------------------------------------------
2[12]

I'll leave it to you to simplify this.

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Polyhedra

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