Optimization: Minimum Area

Date: 11/07/97 at 03:04:44
From: kiyoshi yamashita
Subject: Optimization

A piece of paper (rect. with width a and unlimited length) is folded 
so one corner just reaches the righthand side. How do you fold to make 
the area of the folded paper a minimum.  

I've tried using calculus and derivatives and even just finding a 
function for the area of the triangle. I can't, however, find a 
relation between the sides.  I've tried the law of sines and the 
Pythagorean theorem. I ended up with the fact that the two legs are 
always equal but I know that's not true.

Here's some of my work:

Given: right triangle with Legs x,y and hyp. z; angleYZ=theta; and 
area A.

         A = .5xy
       z^2 = x^2+y^2
cos(theta) = x/y.

     sin(theta)/x = sin(90)/z ---> x = z*sin(theta)
              z^2 = z^2*sin^2(theta) + y^2
z^2(cos^2(theta)) = y^2
    z^2*(x^2/z^2) = y^2
              x^2 = y^2
                x = y

therefore:  A = .5x^2.  

I've also worked to the fact that   A = .5*x*z*cos(theta)

My final attempt included the fact that tan(theta) = x/y.

Thank you so much.

Kiyoshi

Date: 11/07/97 at 08:53:08
From: Doctor Anthony
Subject: Re: Optimization

If you take a length x along the width, and fold it, leaving (a-x) 
unfolded, and let the corner meet the righthand edge somewhere along 
its length such that the side x makes an angle (theta) with the 
length, then we can calculate the area of the folded triangle in terms 
of a and theta.

The other edge of the folded portion (at right angles to x) will make 
an angle theta with the width, and it is easy to see that this length 
is given by a/cos(theta)

We also have   x.sin(theta) = a-x

            x(1+sin(theta)) = a

                                  a
                          x = -----------
                               1 + sin(theta)

The area of the folded triangle = (1/2)x.a/cos(theta)

                                        a^2
                        A   =  ------------------------
                               2cos(theta)[1+sin(theta)]

                                        a^2    
                            =   -------------------------
                                2cos(theta) + sin(2.theta)

dA/d(theta) = 

   a^2[2cos(theta)+sin(2.theta)] x 0 - (-2sin(theta) + 2cos(2.theta)]  
  -------------------------------------------------------------------
                   [2cos(theta) + sin(2.theta)]^2

and putting this equal to 0 we have

                   2sin(theta) - 2cos(2.theta) = 0     

                 sin(theta - (1- 2sin^2(theta) = 0

                2sin^2(theta) + sin(theta) - 1 = 0

              (2sin(theta) -1)(sin(theta) + 1) = 0

and the one practical solution in terms of this problem is:

                                   2sin(theta) = 1

                 sin(theta) = 1/2     so theta = 30 degrees
                
                     
and then    x = a/(1+sin(theta))

            x = a/(1+1/2)

              = a /(3/2)

              = 2a/3

So to get the minimum area you must fold a length 2/3 of 'a' over to 
meet the righthand long edge of the paper.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 11/07/97 at 12:59:42
From: Kiyoshi Yamashita
Subject: Re: Optimization

Thank you, thank you, thank you!