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Math Help - Maximum area of a Triangle

  1. #1
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    Maximum area of a Triangle

    Suppose a triangle has perimeter 2. Let
    x and y be the lengths of two of its sides and A its area. Show that A^2 = (x + y 1)(1 x)(1 y). Use this formula to find the maximum area of a triangle with perimeter 2.

    Hint: You will need to solve a maximization problem

    over a closed, bounded set
    D. To determine D, use the triangle inequality.

    Kind of lost with where to begin. Any help and/or suggestions would be great! Thanks
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JoAdams5000 View Post
    Suppose a triangle has perimeter 2. Let
    x and y be the lengths of two of its sides and A its area. Show that A^2 = (x + y 1)(1 x)(1 y). Use this formula to find the maximum area of a triangle with perimeter 2.

    Hint: You will need to solve a maximization problem

    over a closed, bounded set
    D. To determine D, use the triangle inequality.

    Kind of lost with where to begin. Any help and/or suggestions would be great! Thanks

    Use hero's formula to determine A^2
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JoAdams5000 View Post
    Suppose a triangle has perimeter 2. Let
    x and y be the lengths of two of its sides and A its area. Show that A^2 = (x + y 1)(1 x)(1 y). Use this formula to find the maximum area of a triangle with perimeter 2.

    Hint: You will need to solve a maximization problem

    over a closed, bounded set
    D. To determine D, use the triangle inequality.

    Kind of lost with where to begin. Any help and/or suggestions would be great! Thanks
    and to find Maximum value of A^2, you can use AM-GM inequality.
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  4. #4
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    Use Heron's formula :

     s = 2/2 = 1 and the other side is
     2 - x - y

    Therefore ,

     A = \sqrt{(s)(s-x)(s-y)[s-(2-x-y)]}

     A^2 = (1)(1-x)(1-y)(1-2+x+y) = (1-x)(1-y)(x+y-1)

    Consider

     \frac{ (1-x) + (1-y) + (x+y-1) }{3} \geq [(1-x)(1-y)(x+y-1)]^{\frac{1}{3}}

     (\frac{1}{3})^3 \geq A^2

     A \leq \sqrt{\frac{1}{27}} = \frac{ \sqrt{3} }{9}

    so the maximum value of A is   \frac{ \sqrt{3} }{9}

    The equality holds when

     (1-x) = (1-y) = (x+y-1)

    it gives  x=y= \frac{2}{3} and the other side is also \frac{2}{3} . It is an equilateral  \Delta .
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