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Maximum Volume: Making a Box from a Sheet of Paper

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Date: 10/21/1999 at 08:37:02
From: Chris Leahy
Subject: Differentiation

Hi,

I'm working on a very important question that involves determining the
largest possible volume when making a box out of a sheet of paper. To
do this, you have to cut out squares in the corners of the paper. When
using a square, I noticed that the side of the square equals a sixth
of the size of the length of the paper. I only worked this out by
noticing patterns in several tables that I drew up. I asked a friend
of mine and apparently it has something to do with differentiation. I
terms to me so that I can explain how to show that the length of the
side of the square equals a sixth of the length of a side of the piece
of paper?

I would be most grateful if you could help me with this.

Kind regards,
Chris Leahy
```

```
Date: 10/21/1999 at 21:38:27
From: Doctor Jeremiah
Subject: Re: Differentiation

Differentiation is an algebra method to calculate the slope of a graph
without graphing it.

In your case we will say that the width of the piece of paper is W. I
am also going to assume that we want an open box (no lid). The value
you want to find is x.

Then we get a diagram that looks like this:

|<--------------W-------------->
|<--x-->|<--A = W-2x-->|<--x-->|

+-------+--------------+-------+  ---------
|       |              |       |   |     |
|       |              |       |   |     x
|       |              |       |   |     |
+-------+--------------+-------+   |    ---
|       |              |       |   |     |
|       |              |       |   |     |
|       |              |       |   W   B = W-2x
|       |              |       |   |     |
|       |              |       |   |     |
+-------+--------------+-------+   |    ---
|       |              |       |   |     |
|       |              |       |   |     x
|       |              |       |   |     |
+-------+--------------+-------+  ---------

The volume of the box is A*B*x:

V = A*B*h

= (W-2x)(W-2x)x

= (W^2-4Wx+4x^2)x

= W^2x-4Wx^2+4x^3

The graph of this equation will have a maximum. This is the value we
are looking for. But we don't want to graph the equation and look
for the maximum.

Maximums and minimums are found when the slope is zero. When the slope
is zero the graph is changing direction to go back the way it came and
that is a maximum or minimum.

If we differentiate this equation we get an equation of the graph's
slope. And wherever that equation is equal to zero is a maximum or
minimum.

The differentiation operator looks like d/dx() so d/dx(V) means the
differential of V. The differential of V is the slope of the graph of
V.

V = W^2x-4Wx^2+4x^3

d/dx(V) = d/dx(W^2x-4Wx^2+4x^3)

= d/dx(W^2x) - d/dx(4Wx^2) + d/dx(4x^3)

= W^2*d/dx(x) - 4W*d/dx(x^2) + 4*d/dx(x^3)

= W^2*1 - 4W*2x + 4*3x^2

= W^2 - 8Wx + 12x^2

Remember that d/dx(V) is the slope and the maximum x is when the slope
is 0, so:

d/dx(V) = W^2-8Wx+12x^2

0 = W^2-8Wx+12x^2

Now we have to use the quadratic formula to find the values for the
maximum and minimum values of x.

max/min x = (-b +- sqrt(b^2-4ac))/(2a)

= (8W +- sqrt((-8W)^2 - 4(12)(W^2)))/(2*12)

= (8W +- sqrt(64W^2 - 48W^2))/24

= (8W +- sqrt(16W^2))/24

= (8W +- 4W)/24

= 12W/24 and 4W/24

= W/2 and W/6

The x = W/2 answer gives us:

V = W^2x-4Wx^2+4x^3

= W^2(W/2) - 4W(W/2)^2 + 4(W/2)^3

= W^3/2 - 4W(W^2/4) + 4(W^3/8)

= W^3/2 - 4W^3/4 + 4W^3/8

= W^3/2 - W^3 + W^3/2

= W^3 - W^3

= 0

So the x = W/2 answer is the minimum volume and the x = W/6 answer
must be the maximum volume.

V = W^2(W/6) - 4W(W/6)^2 + 4(W/6)^3

= W^3/6 - 4W(W^2/36) + 4(W^3/216)

= W^3/6 - 4W^3/36 + 4W^3/216

= W^3/6 - W^3/9 + W^3/54

= 9W^3/54 - 6W^3/54 + W^3/54

= 4W^3/54

= 2W^3/27

The maximum volume (2W^3/27) of the box is found when the side of the
box is W/6.

If you need a better explanation, please write back.

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

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