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Math Help - Pentagon Problem

  1. #1
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    Pentagon Problem

    Pentagon Problem-tr.jpg

    Pentagon ABCDE

    BC = CD = BD = AE

    Perimeter (ABCDE) = 10

    Find the sides of rectangle, for which the area of the pentagon will be maximum.

    Please help! i need it so much

    (sry for my english)
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  2. #2
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    Re: Pentagon Problem

    Let BC = x and AB = y. The area of triangle BCD is \frac{x^2 \sqrt{3}}{4}, and the area of ABDE is xy. Hence the area of the pentagon P is


    [P] = \frac{x^2 \sqrt{3}}{4} + xy = \frac{x^2 \sqrt{3} + 4xy}{4}


    However you know that the perimeter is 10, so 3x + 2y = 10 \Rightarrow y = \frac{10-3x}{2}. Substitute into the area equation to obtain


    [P] = \frac{x^2 \sqrt{3} + 4x(\frac{10-3x}{2}))}{4}


    Simplify, and differentiate both sides with respect to x and find critical points.
    Last edited by richard1234; June 19th 2012 at 12:24 PM.
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  3. #3
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    Re: Pentagon Problem

    the answer is 10/(6-sqr(3)) and (15-5sqr(3))/(6-sqr(3)) but i cant get those answers
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  4. #4
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    Re: Pentagon Problem

    Quote Originally Posted by Telo View Post
    Find the sides of rectangle, for which the area of the pentagon will be maximum.

    Please help! i need it so much
    Suppose that the vertical sides of the rectangle each have length a, and suppose that the other sides have length b.

    The area of the pentagon is

    A = ab + \frac12b\left(\frac{\sqrt3}2b \right)

    \Rightarrow A = ab + \frac{b^2\sqrt3}4,

    and the perimeter is

    P = 2a + 3b = 10

    \Rightarrow b = \frac13\left(10 - 2a\right).

    Substituting this into the area equation above produces

    A = a\left[\frac13(10-2a)\right] + \left[\frac13(10-2a)\right]^2\frac{\sqrt3}4

     = \frac19\left[\left(\sqrt3-6\right)a^2 + \left(30 - 10\sqrt3\right)a + 25\sqrt3\right]

    Differentiating,

    \frac{dA}{da} = \frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]

    We locate the critical value:

    \frac{dA}{da} = 0

    \Rightarrow\frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]=0

    \Rightarrow\left(2\sqrt3 - 12\right)a = 10\sqrt3 - 30

    \Rightarrow a = \frac{10\sqrt3 - 30}{2\sqrt3 - 12} = \frac{5\sqrt3 - 15}{\sqrt3 - 6}

    \Rightarrow a = \frac{15 - 5\sqrt3}{6 - \sqrt3}

    Now you can find b.
    Last edited by Reckoner; June 19th 2012 at 11:25 AM.
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  5. #5
    MHF Contributor Reckoner's Avatar
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    Re: Pentagon Problem

    Quote Originally Posted by richard1234 View Post
    However you know that the perimeter is 10, so 4x + 2y = 10 \Rightarrow y = 5-2x.
    Small correction: that should be 3x + 2y = 10. We want the perimeter of the pentagon, so we just need the outer sides.
    Thanks from richard1234
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  6. #6
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    Re: Pentagon Problem

    thanks alot!
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  7. #7
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    Re: Pentagon Problem

    @Reckoner whoops. Can't do math in my head anymore lol. I just fixed my original post.
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