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Optimization of a folded piece of paper

  1. Oct 28, 2009 #1
    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. jcsd
  3. Oct 28, 2009 #2

    lanedance

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    Homework Helper

    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
    [tex] x^2+y^2=(8.5-y)^2 [/tex]

    let 8.5 = c if it makes it easier
    [tex] x^2+y^2=(c-y)^2 [/tex]
     
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