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Math Help - show that area =

  1. #1
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    show that area =

    Hi All,

    OP and OQ are radii of length r cm of a circle centred at O.
    The arc PQ of the circle subtends an angle of theta radians at O and the perimeter of the sector OPQ is 12 cm.

    Show that the are A cm squared of the sector is given by

    A = 72 theta / (2 + theta)^2

    Any help would be appreciated
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Joel View Post
    Hi All,

    OP and OQ are radii of length r cm of a circle centred at O.
    The arc PQ of the circle subtends an angle of theta radians at O and the perimeter of the sector OPQ is 12 cm.

    Show that the are A cm squared of the sector is given by

    A = 72 theta / (2 + theta)^2

    Any help would be appreciated
    r+r+r\theta=12

    2r+r\theta=12

    r=\frac{12}{2+\theta}

    A=\frac{1}{2}{r^2}{\theta}

    =\frac{1}{2}(\frac{12}{2+\theta})^2*\theta

    =\frac{72\theta}{(2+\theta)^2}
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  3. #3
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    Thanks,

    I am now asked to find the maximumarea of this sector. Answer in the book gives me 9cm^2.

    How do I show this??? Do I differentiate or am I on the wrong path?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Joel View Post
    Thanks,

    I am now asked to find the maximumarea of this sector. Answer in the book gives me 9cm^2.

    How do I show this??? Do I differentiate or am I on the wrong path?
    For maximum area of the sector, \frac{dA}{d\theta}=0.
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  5. #5
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    to find dy / dx of (72 theta) / (2 + theta)^2

    do i get rid of the fraction by multiplying 72 theta by (2 + theta)^2??

    This may be completly wrong

    I get 288theta + 288theta^2 + 78 theta^3

    dy dx = 576theta + 234theta

    I think this is just flat out WRONG.

    Any help appreciated
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  6. #6
    MHF Contributor alexmahone's Avatar
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    \frac{dA}{d\theta}=\frac{(2+\theta)^2*72-72\theta*2(2+\theta)}{(2+\theta)^4}
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  7. #7
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    to make this = 0

    2 + theta would have to be 0 = theta = -2

    is this correct?
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  8. #8
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Joel View Post


    to make this = 0

    2 + theta would have to be 0 = theta = -2

    is this correct?
    Incorrect!

    Factorise the numerator and equate it to zero.
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  9. #9
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    thanks heaps for your patience,

    factorise the numerator,

    [IMG]file:///C:/DOCUME%7E1/Nate/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]

    (2 + theta)^2 . 72(1 - theta) . 2(2 + theta)

    am I on the right track ?
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  10. #10
    MHF Contributor alexmahone's Avatar
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    \frac{dA}{d\theta}=\frac{(2+\theta)^2*72-72\theta*2(2+\theta)}{(2+\theta)^4}

    = \frac{72}{(2+\theta)^4}\{(2+\theta)^2-2\theta(2+\theta)\}

    = \frac{72}{(2+\theta)^4}\{(2+\theta)(2+\theta-2\theta)\}

    = \frac{72}{(2+\theta)^3}(2-\theta)

    \frac{dA}{d\theta}=0\implies\theta=2
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  11. #11
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    so putting 2 back into equation



    = 144 / 16

    = 9

    Thats the answer.... woo hoo.

    Thanks very much for your patience and guidance. Very much appreciated.
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