Maximize area of triangle from folded 8.5 by 11 inch paper

[doug1122]

Homework Statement

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 ). Find the value of x that makes the area of the right triangle A a maximum?

Homework Equations

A = 1/2(xy)
x²+y²=(8.5-y)²

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.

 

[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
$$x^2+y^2=(c-y)^2$$