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Math Help - An Applied Trig/Max and Min Problem

  1. #1
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    [SOLVED] An Applied Trig/Max and Min Problem

    Hi there,

    I'm currently struggling with the following question from my textbook.

    "Points A and B lie on a circle, centre O, radius 5 cm. Find the value of angle B that produces a maximum area for triangle AOB."

    I know that I need to find the area and then the derivative and set it to zero but I'm not sure how to find the base and height of the triangle to get the area.

    Thanks!
    Last edited by newslang; January 10th 2012 at 10:47 PM. Reason: SOLVED
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: An Applied Trig/Max and Min Problem

    Use the formula for the area A of a triangle:

    A=\frac{1}{2}ab\sin\theta

    You don't even need calculus to find the angle that maximizes A...
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  3. #3
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    Re: An Applied Trig/Max and Min Problem

    Okay, right. I was just over thinking it.

    So then, it's just \pi/2?

    Thanks again MarkFL
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: An Applied Trig/Max and Min Problem

    you got it!
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  5. #5
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    Re: An Applied Trig/Max and Min Problem

    Quote Originally Posted by MarkFL2 View Post
    you got it!
    Except that cannot form a triangle with both A and B on the circle and andgle OBA=90 degrees!.

    Draw a picture!

    CB
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  6. #6
    Grand Panjandrum
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    Re: An Applied Trig/Max and Min Problem

    Quote Originally Posted by MarkFL2 View Post
    Use the formula for the area A of a triangle:

    A=\frac{1}{2}ab\sin\theta

    You don't even need calculus to find the angle that maximizes A...
    What are a and b, how are they related to the angle OBA?

    Also do not use the same symbol for two different thing in the same problem: A - area of and label of a vertex of the triangle.

    CB
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: An Applied Trig/Max and Min Problem

    My apologies for my sloppiness.

    Using coordinate geometry:

    Putting point O at (0,0), point A at (5,0) and B at \left(5\cos\theta,5\sin\theta\right) where \theta=\angle AOB then the area T of the triangle is:

    T=\frac{1}{2}(OA)(OB)\sin(\theta)
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  8. #8
    Grand Panjandrum
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    Re: An Applied Trig/Max and Min Problem

    Quote Originally Posted by MarkFL2 View Post
    My apologies for my sloppiness.

    Using coordinate geometry:

    Putting point O at (0,0), point A at (5,0) and B at \left(5\cos\theta,5\sin\theta\right) where \theta=\angle AOB then the area T of the triangle is:

    T=\frac{1}{2}(OA)(OB)\sin(\theta)
    OK the area is maximised when your \theta=\pi/2 , but that is not what was asked for, what was asked for is angle OBA (which is now trivial to find, but your original presentation cannot have been anything but confusing to the OP).


    CB
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