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  1. #1
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    maximum area

    An isoceles triangle has two sides of length (l), and the angle between the sides has measure theta. Determine the angle theta that maximizes the area of the triangle, and find the maximum possible area.maximum area-imag0263.jpg
    I know the answer is pi/2 but i dont know how? Please help!!
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    Quote Originally Posted by purplelollipop27 View Post
    An isoceles triangle has two sides of length (l), and the angle between the sides has measure theta. Determine the angle theta that maximizes the area of the triangle, and find the maximum possible area.Click image for larger version. 

Name:	IMAG0263.jpg 
Views:	159 
Size:	377.1 KB 
ID:	19979
    I know the answer is pi/2 but i dont know how? Please help!!
    for any isosceles triangle with equal sides L and vertex angle \theta ...

    A = \frac{1}{2} L^2 \sin{\theta}
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  3. #3
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    \displaystyle A=\frac{1}{2}bh

    Consider 3 possibilities:

    \displaystyle \frac{\theta}{2}+\frac{\theta}{2}=\theta

    That assume the angle isn't split at first to solve the whole area of the triangle.

    Theta =90
    Theta < 90
    Theta > 90
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  4. #4
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    Maximize area of isosceles triangle.

    Quote Originally Posted by purplelollipop27 View Post
    An isosceles triangle has two sides of length (l), and the angle between the sides has measure theta. Determine the angle theta that maximizes the area of the triangle, and find the maximum possible area.Click image for larger version. 

Name:	IMAG0263.jpg 
Views:	159 
Size:	377.1 KB 
ID:	19979
    I know the answer is pi/2 but i don't know how? Please help!!

    Modify your attached figure so that the equal sides have length, \ell, rather than 1. Then the altitude, \displaystyle h, is given by \displaystyle h= \ell\cos{\theta\over2} and the base, \displaystyle b, is given by \displaystyle b=2\ell\sin{\theta\over2}

    The area is: \displaystyle A={1\over2}bh={1\over2}\left(2\ell\sin{\theta\over  2}\right)\left(\ell\cos{\theta\over2}\right)

    Use \displaystyle \sin(2x)=2\sin(x)\cos(x), with \displaystyle x={\theta\over2} to simplify this to

    \displaystyle A={1\over2}\,\ell\, ^2\sin\theta.

    Maximize the area with respect to \displaystyle\theta.

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