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Quadriateral Max/Min Question

  1. Jun 2, 2007 #1
    1. The problem statement, all variables and given/known data

    The diagonals of a quadrilateral are perpendicular. The sum of the lengths of the diagonals is 6cm. What is the maximum area of such a quadrilateral?

    2. Relevant equations



    3. The attempt at a solution

    let x and y be the lengthes of the diagonals. Then the area of the quadrilateral is calculated by:

    a = ½ * x * y

    From the given conditions you know: x + y = 6 ==> y = 6 - x

    Plug in the term for y into the first equation:

    a(x) = ½*x*(6 - x) = -½x² + 3x


    I do not know what to do from here. I know that there is a graphing method but I'd rather do it through differentiation so could someone do the solution that way. The answer should be y=3 and x=3 so the dimensions are 3 x 3 I think.
     
  2. jcsd
  3. Jun 2, 2007 #2
    Do you know how to find the extrema(maxima and minima) of a given function with respect to a variable?
     
  4. Jun 2, 2007 #3
    I'm more confused now. If I differentiate a(x), this is what happens
    a'(x)=2/2 x + 3
    0= x + 3
    x= -3

    y= 6 - x
    y = 6 - (-3)
    y= 9

    So the dimensions are 9 x 3? That doesn't seem right.
     
  5. Jun 2, 2007 #4
    You missed the minus sign.
     
  6. Jun 2, 2007 #5
    6- -3 is 9.

    I'm going over it and that's still the only answer I get for some reason.
     
  7. Jun 2, 2007 #6
    I meant when you differentiated the original equation.
     
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