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Creasing Paper


Date: 05/11/97 at 14:17:14
From: Becky & Tommi
Subject: The shortest crease (minimizing)

Dear Dr. Math,

We are in a college credit calculus class and we are doing our
quarter project. The problem asks us to find the shortest crease when 
folding a piece of paper 6 units wide by 25 units long. We are to fold 
the lower left corner to touch the right side, so that the crease will 
be minimized.

We have tried several different ideas, but the one that gave us the
shortest crease was when we folded the loewr left corner to the upper 
right corner. How do we explain this using calculus? We understand 
how to take the derivative and set it eqaul to 0 to find the minimum, 
but how do we relate that to this problem?
 
Can you help?

Thanks,
Becky and Tommi 


Date: 05/12/97 at 08:04:09
From: Doctor Anthony
Subject: Re: The shortest crease (minimizing)

Dear Becky and Tommi,

Let x = length along the bottom edge from the righthand corner to 
where the fold starts. After folding the bottom lefthand corner to 
meet the righthand edge of the paper, we have a right triangle with 
bottom length x, hypotenuse (6-x) and, using the Pythagorean theorem, 
the vertical side given by:
 
  sqrt((6-x)^2 - x^2)) = sqrt(36-12x)

Now if you fold the paper back to its original position as a flat, 
rectangular sheet, you can calculate the distance that the bottom 
lefthand corner has to move to reach the right hand edge. This 
distance is the hypotenuse of a right triangle having a bottom side of 
6 units and a vertical side sqrt(36-12x). The length of the hypotenuse 
is sqrt(36 + 36-12x).

  Length of hypotenuse = sqrt(72-12x)

Now fold the corner up again to meet the righthand edge and draw a 
perpendicuar from this displaced corner to meet the fold in the paper.  
The length of this perpendicular is half the length of the hypotenuse 
just calculated: (1/2)sqrt(72-12x).

Using the Pythagorean theorem, the distance from where the fold starts 
on the bottom edge of the paper to the foot of the perpendicular 
described above is given by:
          
          sqrt((6-x)^2 - (1/4)(72-12x))
        = sqrt(36-12x+x^2 - 18+3x)
        = sqrt(18-9x+x^2)

We now have the information necessary to calculate, in terms of x, the 
length of the fold.

If L = length of fold, we have by similar triangles:

   L/(6-x) = (6-x)/sqrt(18-9x+x^2)

         L = (6-x)^2/sqrt(18-9x+x^2)

We differentiate and equate to zero to find the value of x giving the 
minimum length of fold:

     sqrt(18-9x+x^2)2(6-x)(-1) - (6-x)^2(1/2)(18-9x+x^2)^(-1/2)(-9+2x)
dL/dx=----------------------------------------------------------------
                     18 - 9x + x^2


Only the numerator need concern us, if this expression is put equal 
to 0.

Divide out by (6-x):

                           (1/2)(6-x)(-9+2x)
      -2sqrt(18-9x+x^2) - -----------------    = 0
                           sqrt(18-9x+x^2)

           -2(18-9x+x^2) = (1/2)(6-x)(-9+2x)
           -4(18-9x+x^2) = -54 + 21x - 2x^2
         -72 +36x - 4x^2 = -54 + 21x - 2x^2
        -2x^2 + 15x - 18 = 0
         2x^2 - 15x + 18 = 0
             (2x-3)(x-6) = 0

So x = 3/2  or  x = 6

Clearly x = 6 is the maximum length, and we require x = 3/2 to give 
the minimum length of fold.

Then L = (6-x)^2/sqrt(18-9x+x^2)    

Putting in x = 3/2:

      L = 7.79423 units

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

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