Optimization of a folded piece of paper

1. Oct 28, 2009

doug1122

1. The problem statement, all variables and given/known data

If you take an 8.5in by 11in piece of paper and fold one corner over so it just touches the opposite edge as seen in figure (http://wearpete.com/myprob.jpg [Broken]). Find the value of x that makes the area of the right triangle A a maximum?

2. Relevant equations
A = 1/2(xy)
x2+y2=(8.5-y)2

3. The attempt at a solution
x2+y2=(8.5-y)2
x = sqrt((8.5-y)2-y2)
A = 1/2(y)(sqrt((8.5-y)2-y2))
da/dx = ((y2-4.25y)/sqrt((8.5-y)2-y2))+1/2(sqrt(-17y-72.25))
I know that at da/dx=0 the triangle is maximized but da/dx is undefined at y=0 (y graphically). I am pretty sure my derivative is right but maybe I missed something there. Thanks for taking a look.

Last edited by a moderator: May 4, 2017
2. Oct 28, 2009

lanedance

at y = 0, the area will be zero, so i wouldn't be too concerned about that point

your method is ok, but could be simplified a bit... try muliplying out the RHS of your equation and simplifyng before substituting in
$$x^2+y^2=(8.5-y)^2$$

let 8.5 = c if it makes it easier
$$x^2+y^2=(c-y)^2$$