The Shortest Crease

Date: 12/29/97 at 03:14:48
From: Kathy
Subject: The shortest crease

Q: A paper which is 6 units one side and 25 units another side. Put 
the shorter side to face you. Fold the lower right corner to the left 
side, making a crease. Our task is to find the minimum crease by 
folding the right corner to the left side. 

HINT: this question is related to derivative - you may use derivative 
to solve it. (You are required to write down the equations that you 
use and explain how you solve it.)

This question is a project given in my calculus class. We are 
supposed to write equations to solve it. We have to write a bunch of 
explanations and equations to complete a project paper. So far, I 
have no idea how to solve it. Could you help me? 

Thank you very much for your help. 
                                                 
Sincerely, 
Kathy

Date: 12/29/97 at 12:22:00
From: Doctor Anthony
Subject: Re: The shortest crease

I have taken a general case with width equal to k, so you can see how 
the working can be adjusted for different widths of paper.

If you take a length x along the width, and fold it, leaving (k-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 length of the fold and the area of 
the folded triangle in terms of k 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 k/cos(theta)

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

               x(1+sin(theta)) = k

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

The length of the fold, by Pythagoras is given by 

  FOLD^2  = x^2 + [k/cos(theta)]^2

          = [k/(1+sin(theta)]^2 + [k/cos(theta)]^2

          = k^2[(1+sin(theta))^(-2) + sec^2(theta)]

Differentiate with respect to theta and we get

 k^2[(-2)(1+sin(theta)^(-3)(cos(theta) + 
2.sec(theta)sec(theta)tan(theta)] 

 k^2[-2.cos(theta)/(1+sin(theta))^3 + 2.sin(theta)/cos^3(theta)] 

and equating this to zero, we get

        cos^4(theta) = sin(theta)(1+sin(theta))^3

putting sin(theta) = 1/3 and then cos(theta) = sqrt(8)/3, this 
equation is satisfied. So the minimum fold occurs with sin(theta) 
= 1/3

    x = k/(1+1/3) =  k/(4/3)  =  (3/4)k
   
  FOLD^2 = (9/16)k^2 + (9/8)k^2  = (27/16)k^2

     Fold = (3.Sqrt(3)/4)k

-Doctor Anthony,  The Math Forum
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