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Math Help - extreme trapeze perimeter

  1. #1
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    extreme trapeze perimeter

    A halfcircle with r = 14 cm has a trapeze drawn inside it.


    What is the largest possible perimeter of the trapeze (check the image).

    I have to express the perimeter using only 1 variable (i was thinking c) and then derivate the perimeter and equal it to 0 to get the extremes..

    how to express the perimeter using only c?
    Attached Thumbnails Attached Thumbnails extreme trapeze perimeter-trapeze.jpg  
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  2. #2
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    ok I tried something..



     b^2 = x \cdot 28

    then the perimeter is equal to:

     o = a + b + c + d = 2r + 2b + c

    c = 28 - 2x
    b = \sqrt{28x}

    therefore:

     o(x) = 28 + 2\sqrt{28x} + 28 - 2x = 56 - 2x + 2\sqrt{28x}

    o'(x) = \frac{1}{2}\cdot 2(28x)^{-\frac{1}{2}} \cdot 28 - 2 = \frac{28}{\sqrt{28x}} -2


    to find the extremes:

     o'(x) = 0
    \frac{28}{\sqrt{28x}} -2 = 0

     28 = 2\sqrt{28x} /^2
     784 = 112x
    x = 7

     o(x) = 56 - 2x + 2\sqrt{28x}

     o(7)  = 56 - 2 \cdot 7 + 2\sqrt{28 \cdot 7} = 42 + 28 = 70


    To check wether the extreme is max or min:

     o''(x) = (\frac{28}{\sqrt{28x}} -2)' = 28 \cdot (-\frac{1}{2})\cdot (28x)^{-\frac{3}{2}} \cdot 28 = -\frac {392}{\sqrt{(28x)^3}} (< 0) \Rightarrow Maximum

    can someone check if I'm right please :P
    Attached Thumbnails Attached Thumbnails extreme trapeze perimeter-trapeze.jpg  
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