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Who wants to help me maximize the area of a kite? (Optimization)

  1. Aug 10, 2009 #1
    I just can't figure this problem out.

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

    You have four pieces of wood, two with length a and two with length b, and you arrange them in the shape of a kite (pieces of equal length placed adjacent to each other). You want to build a cross in the middle as a support. How long should the middle pieces be to maximize the area of the kite?

    Sorry, there was a picture in the textbook, but it should be pretty straight forward.

    2. Relevant equations

    It's optimization.

    3. The attempt at a solution

    I can't figure out how to relate the variables to each other. I know that you can cut the kite in half symmetrically, and rearrange the kite into a rectangle with length y and width x. That was what I tried to do, but I still couldn't figure it out. I ended up with a whole bunch of square roots and I couldn't isolate the variable.

    I tried arranging the pieces in different triangles, but I ended up with too many variables.

    Any hints would be appreciated.
     
  2. jcsd
  3. Aug 11, 2009 #2

    Mark44

    Staff: Mentor

    Are you sure you have posted the information of this problem exactly as given? For one thing, you don't make a kite by building a frame around it. And besides, after you have used your four pieces of wood, what are you supposed to use to build the cross pieces?

    For another thing, if a != b, your kite is in the shape of a rhombus; the angle formed by a short piece and a long piece is not 90 degrees. You can't cut the kite in half and form a rectangle. Instead, what you get is a parallelogram.
     
  4. Aug 11, 2009 #3
    sorry, i meant cut the kite into four. then you can rearrange it into a rectangle.

    i was looking at the problem in the textbook and i noticed a little symbol beside the number. it said "CAS". and so i searched the textbook for what "CAS" meant and it turns out it stands for "computer algebra system". in other words, my instructor assigned us a problem that she shouldn't have, because we don't learn how to use computer algebra systems in this course. thanks anyway.
     
  5. Aug 11, 2009 #4

    Mark44

    Staff: Mentor

    No you can't, not of the values of a and b are different. For a rectangle, sides a and b have to meet at a 90 degree angle. This can't happen if a and b are different.
     
  6. Aug 11, 2009 #5
    kitetorectangle.png
     
  7. Aug 11, 2009 #6

    Mark44

    Staff: Mentor

    I stand corrected.
     
  8. Aug 11, 2009 #7
    hehe

    either way, the problem is too hard for the level i'm at. need a computer.
     
  9. Aug 11, 2009 #8
    [tex]P = \frac{1}{2}d_1 d_2 = ab \sin(\angle ab)[/tex]

    If we represent is as function f(x)=absin(x), then to maximize the area f'(x)=0 and f''(x)>0.

    So f'(x)=abcos(x) .

    abcos(x)=0

    cos(x)=0

    And x=п/2.

    f''(x)=absin(x)

    f''(п/2)=ab>0 and for x=п/2 the area is maximized.
     
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