## Wednesday, April 8, 2015

### Minimize The Crease

Consider a rectangular piece of paper of width $W$ laid on a flat surface. The lower left corner of the paper is bought over to the right edge of the paper, and the paper is smoothed flat creating a crease of length $L$, as in the diagram:

What is the minimal value of $L$ in terms of $W$?

I have filled in the previous diagram with the information I need:

By similarity, we may state:

[MATH]\frac{\sqrt{L^2-x^2}}{W}=\frac{x}{\sqrt{W(2x-W)}}[/MATH]

Squaring, we obtain:

[MATH]\frac{L^2-x^2}{W^2}=\frac{x^2}{W(2x-W)}[/MATH]

[MATH]L^2-x^2=\frac{Wx^2}{2x-W}[/MATH]

[MATH]L^2=\frac{2x^3}{2x-W}[/MATH]

At this point we see that we require [MATH]\frac{W}{2}[/MATH].

Minimizing $L^2$ will also minimize $L$, and so differentiating with respect to $x$ and equating to zero, we find:

[MATH]\frac{d}{dx}\left(L^2 \right)=\frac{(2x-W)\left(6x^2 \right)-(2)\left(2x^3 \right)}{(2x-W)^2}=\frac{2x^2(4x-3W)}{(2x-W)^2}=0[/MATH]

Discarding the root outside of the meaningful domain, we are left with:

[MATH]4x-3W=0[/MATH]

[MATH]x=\frac{3}{4}W[/MATH]

The first derivative test easily shows that this is a minimum, as the linear factor in the numerator, the only factor which changes sign, goes from negative to positive across this critical value.

Thus, we may state:

[MATH]L_{\min}=L\left(\frac{3}{4}W \right)=\sqrt{\frac{2\left(\frac{3}{4}W \right)^3}{2\left(\frac{3}{4}W \right)-W}}=\frac{3\sqrt{3}}{4}W[/MATH]