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Homework Help: Window area question, express as function of Area

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data

    A window has the shape of a rectangle surmounted by an equilateral triangle. Given that the perimeter of the window is 15 feet, express the area as a function of the length of one side of the equilateral triangle.

    2. Relevant equations

    Area of an equaliteral triangle : x^2(sqrt (3)) / 4

    Surface area of the window : 3x + 2y = 15
    reduced to : y = (15 - 3x) / 2

    Volume of the window: X^2(sqrt(3)) / 4 + xy

    3. The attempt at a solution

    y = (15 - 3x) / 2 has domain of 0 <= x <= 5

    i subbed in y into the vlume of the window,
    x^2(sqrt(3)) / 4 + x(15-3x)/2

    and after factoring out the x, I got [x(x*Sqrt(3) - 6x + 30)] / 4
    and with new domain 0 < x < 5.


    Am i takin the right approach? am i suppose to leave the sqrt where it is right now?
    please help
     
  2. jcsd
  3. Sep 22, 2010 #2

    hunt_mat

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    Wouldn't the perimeter be 2x+2y=15? Then the area of the window is given by A=xy, but you know that x+y=7.5, then...
     
  4. Sep 22, 2010 #3

    Mentallic

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    No, you're missing the equilateral triangle.

    Lovemake1 yes that's perfect.
     
  5. Sep 22, 2010 #4

    hunt_mat

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    No, it says the perimeter of the 15, if the triangle fits snugly into the rectangle and the sides of the triangle is L, the one side is length L and the other side is given by [tex]L\sqrt{3}/2[/tex], so the area is given by [tex]L^{2}\sqrt{3}/2[/tex].
     
  6. Sep 22, 2010 #5

    Mentallic

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    Surmounted means to sit on top of, not to sit snugly into.

    And by the way, for an equilateral triangle, if one side is length L then the area is [tex]\frac{\sqrt{3}}{4}L^2[/tex]
     
  7. Sep 22, 2010 #6

    hunt_mat

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    Ah! I read surrounded.
     
  8. Sep 22, 2010 #7

    Mentallic

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    That would make for a weird question :tongue:
     
  9. Sep 22, 2010 #8

    hunt_mat

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    Not quite, it would mean that you could get a number for the area by using the perimeter.

    Mat
     
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