Calculus word problem: folded paper

**Bowlbase** · September 29th 2011, 07:18 PM

A very long rectangular piece of paper is 20 cm wide. The bottom right hand corner is folded along the crease so that the corner just touches the left hand side of the page. How should the page be folded to make the crease as short as possible?

I know that I must use Pythagorean theorem. But, since this is in the trig function chapter I can assume that I'll probably have to use a trig function.

Okay, 20 cm side is "a" side. Crease side (hypotenuse) is "c." And, the remaining side will be "b." c=sqrt[(20-a)^2+b^2]. I know that the Domain for "a" must be 0<a=<20. I also know that as "a" gets closer to 20 "c" gets smaller. Its all well and good that I know the answer must be 20 but I don't get points for observations. Anyhow, knowing that the answer must be a=20 then I know that the angle created by turning the page will be 90 degrees.

I know all this but I don't know how to make the constraint with this information. How do I turn my objective function into a single variable function?

Thanks for the help.

**tom@ballooncalculus** · September 30th 2011, 02:15 AM

Not sure if my pic is labelled the same as yours. (You did one, yes?) My a > 10, by the way.

Similar triangles to get b in terms of a.

Hope that helps.

Also...

Originally Posted by Bowlbase

I also know that as "a" gets closer to 20 "c" gets smaller. Its all well and good that I know the answer must be 20 but I don't get points for observations. Anyhow, knowing that the answer must be a=20 then I know that the angle created by turning the page will be 90 degrees.
I know all this but...

... shock coming!

**Bowlbase** · October 1st 2011, 05:48 AM

I had 10 as the minimum as well, not sure why I dint put it in as such. I follow everything you did there but not sure how to get "c" from this.

And yeah I did drawings, and folded several sheets of paper. That's how I came to the conclusion that it must be 20 as the crease gets longer when I fold it any other way. But, its possible I'm bad at folding paper I guess.

**earboth** · October 1st 2011, 07:57 AM

Originally Posted by Bowlbase

I had 10 as the minimum as well, not sure why I dint put it in as such. I follow everything you did there but not sure how to get "c" from this.

...

1. I've taken the drawing of tom@ballooncalculus and indicated the similar triangles you should use.

2. Using proportions you'll get:

$\dfrac b{20} = \dfrac a{\sqrt{40a-400}}$

3. Solve for b. Use Pythagorean theorem to determine the length of c:
$a^2+\frac{40a^2}{40a-400}=c^2$

4. Use calculus: Differentiate c wrt a and solve c'(a) = 0 for a. To solve this equation it would be best to use a numerical method, for instance Newton's method. I've got $a \approx 13.32917446$ . Consequently calculate the length of c.

**Bowlbase** · October 1st 2011, 10:59 AM

Sorry, crying baby will take me some time to get back to homework.

**tom@ballooncalculus** · October 1st 2011, 11:00 AM

Originally Posted by earboth

1. I've taken the drawing of tom@ballooncalculus and indicated the similar triangles you should use.

Thanks, but I don't see that, actually. I would put the bright yellow up in the top triangle. (Edit: oh, I see you have!) Then the equal proportions are...

$\frac{b}{20}\ =\ \frac{a}{\sqrt{40 (a - 10)}}$

... and you can substitute an 'a' expression for 'b' in,

$c^2 = a^2 + b^2$

... and differentiate that equation. And then set dc/da to zero, and solve for a.

Originally Posted by Bowlbase

And yeah I did drawings, and folded several sheets of paper. That's how I came to the conclusion that it must be 20 as the crease gets longer when I fold it any other way. But, its possible I'm bad at folding paper I guess.

Yes, it's a really great problem, almost a conjuring trick - you could get someone to put money on the minimum being 20.

Did you actually take measurements of the fold, or just assume that the lower down it reached on the vertical side, the shorter?

**earboth** · October 2nd 2011, 12:56 AM

Originally Posted by tom@ballooncalculus

Thanks, but I don't see that, actually. I would put the bright yellow up in the top triangle. (Edit: oh, I see you have!) Then the equal proportions are...

$\frac{b}{20}\ =\ \frac{a}{\sqrt{40 (a - 10)}}$

... and you can substitute an 'a' expression for 'b' in,

$c^2 = a^2 + b^2$

... and differentiate that equation. And then set dc/da to zero, and solve for a.
1

Originally Posted by earboth

...
4. Use calculus: Differentiate c wrt a and solve c'(a) = 0 for a. To solve this equation it would be best to use a numerical method, for instance Newton's method. I've got $a \approx 13.32917446$ . <-- This is definitely wrong. I can't find my mistake. Here are the correct calculations:

Originally Posted by Bowlbase

...But, its possible I'm bad at folding paper I guess.

1. $a^2 + \left(\frac{20a}{\sqrt{40a-400}}\right)^2=c^2 ~ \implies ~ c^2=\frac{a^3}{a-10}$

2. Using c² instead of c you get

$\frac{d(c^2)}{da} = \frac{2a^2 (a-15)}{(a-10)^2}$

3. The RHS of this equation equals zero if a = 0 or a = 15.

**Bowlbase** · October 2nd 2011, 10:19 AM

Alright, I got it. My problem was the proportion. I guess I need to revisit some of my trig book.

I even measured a 20cm piece of paper and the answers for 20 and 15 were both correct to within 3mm. I was so sure 20 would be correct but didn't really notice how the crease grew on the bottom of the paper I guess.

Thanks a lot for the help!!

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