Associated Topics || Dr. Math Home || Search Dr. Math

### 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!
```
Associated Topics:
College Calculus
College Euclidean Geometry
College Trigonometry
High School Calculus
High School Euclidean/Plane Geometry
High School Trigonometry

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