Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Maximum Volume: Making a Box from a Sheet of Paper


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 
don't know anything about this, so could you explain it in simple 
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

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/