My Math Forum Minimizing A Paper Crease [Word Problem]

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December 13th, 2011, 06:12 PM   #1
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Minimizing A Paper Crease [Word Problem]

Here is the problem:
A rectangular sheet of 8 1/2 by 11-in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L.

(a) Show that $L^2\,=\frac{2x^3}{2x\,-8.5}$
(b) What value of $x$ minimizes $L^2$ ?
(c) What is the minimum value of $L$ ?
There is a visual attached.

[attachment=0:264l7z7r]jb1.jpg[/attachment:264l7z7r]
Attached Images
 jb1.jpg (50.0 KB, 993 views)

December 13th, 2011, 08:57 PM   #2
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Math Focus: Calculus/ODEs
Re: Minimizing A Paper Crease [Word Problem]

a) Let W be the width of the paper. Notice that triangle ARP is congruent with triangle QRP. Draw a line segment parallel to DC from R to segment BC and label the point of intersection as S. Label segment PB as W - x. Observe that angles PQB and RQS are complementary and since both angles are in right triangles, we have that triangles QRS and PQB are similar. So, we may state the following:

$RQ=AR=\sqrt{L^2-x^2}$

$BQ=\sqrt{x^2-$$W-x$$^2}=\sqrt{x^2-$$W^2-2Wx+x^2$$}=\sqrt{2Wx-W^2}$

$\frac{BQ}{x}=\frac{W}{RQ}=\frac{W}{\sqrt{L^2-x^2}}$ giving:

$\frac{\sqrt{2Wx-W^2}}{x}=\frac{W}{\sqrt{L^2-x^2}}$

$\frac{\sqrt{2x-W}}{x}=\sqrt{\frac{W}{L^2-x^2}}$

$\frac{2x-W}{x^2}=\frac{W}{L^2-x^2}$

$L^2-x^2=\frac{Wx^2}{2x-W}$

$L^2=\frac{Wx^2}{2x-W}+x^2$

$L^2=\frac{Wx^2+$$2x-W$$x^2}{2x-W}$

$L^2=\frac{2x^3}{2x-W}$

Note that in order for L to represent a real value or a positive value or a meaningful value, we require $\frac{W}{2}. L = 0 for x = 0, but that is a trivial solution and does not allow for the corner at A to be brought over to edge BC.

b) Implicitly differentiate with respect to x:

$2L\frac{dL}{dx}=\frac{$$2x-W$$$$6x^2$$-$$2x^3$$$$2$$}{$$2x-W$$^2}=\frac{2x^2$$4x-3W$$}{$$2x-W$$^2}$

$\frac{dL}{dx}=\frac{2x^2$$4x-3W$$}{2L$$2x-W$$^2}$

The only extremum in the valid domain for x is found with:

$4x-3W=0$

$x=\frac{3}{4}W$

To ensure we have found a minimum, we may check the sign of the only factor that could change sing in the derivative on either side of the critical value:

$4$$\frac{5}{8}W$$-3W=-\frac{W}{2}$ L is decreasing.

$4$$\frac{7}{8}W$$-3W=\frac{W}{2}$ L is increasing, thus we have a minimum value for L when $x=\frac{3}{4}W$.

c) $L=\sqrt{\frac{2x^3}{2x-W}}=\sqrt{\frac{2$$\frac{3}{4}W$$^3}{2$$\frac{3}{4 }W$$-W}}=\sqrt{\frac{\frac{27}{32}W^3}{\frac{1}{2}W}}=\ frac{3\sqrt{3}}{4}W$

Here is a plot of L for x = 0.5 to 1, letting 1 unit represent W:

[attachment=0:kwsh8c5z]creaseplot.jpg[/attachment:kwsh8c5z]
Attached Images
 creaseplot.jpg (18.2 KB, 971 views)

 December 14th, 2011, 05:25 AM #3 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Re: Minimizing A Paper Crease [Word Problem] Thank you very much for the help.

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