# [SOLVED] Quick optimization question

• June 1st 2009, 02:06 PM
sinewave85
[SOLVED] Quick optimization question
I don't understand why in this example problem the textbook authors chose the values they did for the domain of the the variable. Any clarification would be welcome!

The lower right-hand corner of a piece of paper is folded over to reach the leftmost edge, as shown. If the page is 20 cm wide and 30 cm long, what is the length L of the shortest possible crease?

http://i125.photobucket.com/albums/p...thdraw4637.jpg

The text states that $0 \leq x \leq 15$. I don't understand why it is not $0 \leq x \leq 20$ -- that is what I would have used.

How to you know just looking at the basic problem information that the maximum possible value for x is 15?
• June 2nd 2009, 06:03 AM
Media_Man
The text is wrong
People put a lot of faith in the anti-typo magic of text books. Of course x can be greater than 15. It can be shown that x must be at least 10, otherwise the rightmost corner of the paper won't even reach the leftmost wall at all. In fact, only when x is greater than 11.848 will L terminate on the right of the page instead of the top of the page ( L(x) is a piecewise function). Perhaps the text accidentally gave away the answer, as the minimum value of L(x) is in fact x=15.

Do you need help with the problem itself or just this setup question?
• June 2nd 2009, 08:13 AM
Calculus26
I agree completely with Media Man except I get 11.459 for the point beyond which L lies on the right side of the page??????
• June 2nd 2009, 08:29 AM
Media_Man
Blooper
Yes, Calculus26, you are correct. I got $L(x)=\sqrt{\frac{x^3}{x-10}}$ and tried to set up the inequality $L(x)<30$, which is not the breaking point I was looking for. The correct value of x is the solution to $x^2-90x+900=0$, which is indeed 11.45898034.

sinewave85: Again, do you seek assistance on the actual problem, or did you simply need to clear up this one simple text mistake?
• June 2nd 2009, 10:29 AM
sinewave85
You guys pwn the text!
Thanks for the help, Media_Man and Calculus 26. I could not get to my computer this morning, so forgive my tardy reply. This is an example problem is a review book that accompanies my text book -- I have it fully worked in front of me -- and I was just looking for clarification on that one point. I don't have enough confidence in my own mathematical abilities to assume that if I disagree with something in a text book that the point of error is the book and not me, so it is nice to have some reenforcement. Thanks again, Media_Man and Calculus 26!
• June 2nd 2009, 10:33 AM
Calculus26
sinewave---the confidence will come.

I have had seveal students who always thought the answer in the book
was wrong whenever their answer didn't match.

Believe me you're in a better position doubting yourself a little.
• June 2nd 2009, 10:51 AM
sinewave85
The benefits of doubt.
Quote:

Originally Posted by Calculus26
sinewave---the confidence will come.

I have had seveal students who always thought the answer in the book
was wrong whenever their answer didn't match.

Believe me you're in a better position doubting yourself a little.

I think that I was a bit more like that when I was younger, but with age has come the wisdom to recognize my own limitations.

P.S. As a self-taught math student, I can say with confidence that the world needs all of the good math teachers it can get -- I would not recommend that anyone do it the way I have. I hope your students appreciate you!