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Math Help - Maxima och trig equation

  1. #1
    Senior Member Twig's Avatar
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    Maxima och trig equation

    Hi

    The goal is to find the angle x that gives the largest area.
    I have an area expression:

     A(x)= 2L^{2}\cdot sin(\frac{x}{2})+L^{2}\cdot cos(\frac{x}{2}) \cdot sin(\frac{x}{2}) where L is a constant.

    If the derivative below is wrong, let me know. I did use some trig identities.

    A'(x)=L^{2}(cos(\frac{x}{2})+\frac{1}{2}\cdot cos(x))

    So basically, how do I find the x the gives a maxima?

    Thanks
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  2. #2
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    Hello, Twig!

    The goal is to find the angle x that gives the largest area.

    I have an area expression:

     A(x)\:=\: 2L^{2}\sin(\tfrac{x}{2})+L^{2}\cos(\tfrac{x}{2})\s  in(\tfrac{x}{2}) .where L is a constant.


    If the derivative below is wrong, let me know. I did use some trig identities.

    A'(x)\:=\:L^2\left(\cos(\tfrac{x}{2})+\tfrac{1}{2}  \cos(x)\right) .
    This is correct, but . . .

    So basically, how do I find the x the gives a maxima?

    Our equation was: . A'(x) \;=\;L^2\left(\cos\tfrac{x}{2} + \tfrac{1}{2}\cos^2\!\tfrac{x}{2} - \tfrac{1}{2}\sin^2\!\tfrac{x}{2}\right) \;=\;0


    Multiply by \frac{2}{L^2}\!:\quad 2\cos\tfrac{x}{2} + \cos^2\!\tfrac{x}{2} - \sin^2\!\tfrac{x}{2} \;=\;0


    Replace \sin^2\!\tfrac{x}{2} with 1 - \cos^2\!\tfrac{x}{2}\!:\quad 2\cos\tfrac{x}{2} + \cos^2\!\tfrac{x}{2} - (1 - \cos^2\!\tfrac{x}{2}) \;=\;0

    . . And we have: . 2\cos^2\!\tfrac{x}{2} + 2\cos\tfrac{x}{2} - 1 \;=\;0 . . . a quadratic in \cos\frac{x}{2}

    Quadratic Formula: . \cos\tfrac{x}{2} \;=\;\frac{-2 \pm\sqrt{12}}{4} \;=\;\frac{-1\pm\sqrt{3}}{2}


    We find that: . \cos\frac{x}{2} \:=\:\frac{-1 - \sqrt{3}}{2} \:=\:-1.36\hdots . has no real solutions.

    And that: . \cos\frac{x}{2} \:=\:\frac{-1+\sqrt{3}}{2} \quad\Rightarrow\quad \frac{x}{2} \:=\:68.529857^o \quad\Rightarrow\quad x \:=\:137.0585971^o


    . . Therefore: . x \;\approx\;137^o

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  3. #3
    Senior Member Twig's Avatar
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    Hi!

    Thank you so much!

    This approach was much better.

    I tried to re-write with half-angle formulas and stuff, I thought it should work but it didnīt. =)
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