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The Broad Side of the Barn

  1. Jul 19, 2006 #1
    Our teacher gave us this word problem:

    I tried to solve it but got stuck many times - any kind of help will be greatly appereciated.

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 19, 2006 #2
    The thing I would like to know is if you are supposed to know the equation of the 30 *45 wall. It seems like a parabola but is it ?
    If you don't have the equation of the roof's curve, I really don't see how you can proceed analyticaly.
  4. Jul 19, 2006 #3


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    Well, setting up a coordinate system with (0,0) at the center base of the end wall, The roof is a parabola (I guess) with vertex at (0, 45), passing through (-15,30) and (15,30). That tells us that the parabola is symmetric about x=0 and so is of the form y= ax2+ 45. Setting x= 15 and y= 30, we get 30= 225a+ 45 so a= -15/225= -1/15. Once you know that you can calculate the area of that side of the barn.
    The area of the sides, since they are rectangles, is easy. Since you can buy the siding at any length, there should be no waste there. Calculating the waste on the endwalls, once you are into the parabolic section, you can buy the length that reaches the end of the wall at the bottom and the waste is the part that reaches outside the parabola above that. You can reduce that by using the narrowest ( 1 1/2 feet wide) siding.

    (How is this connected with "Riemann sums"?)
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