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Math Help - Optimization parabola/triangle

  1. #1
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    Optimization parabola/triangle

    An isoceles triangle with vertices (0,0), (x,y) & (-x,y) is inscribed in the parabola y=4-x^2

    What x & y will give triangle max area and what is that area?

    I'm unsure of my first move to get an A(x).

    Edit: also to I need to split my triangle into halves for this?
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  2. #2
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    Quote Originally Posted by mattc View Post
    An isoceles triangle with vertices (0,0), (x,y) & (-x,y) is inscribed in the parabola y=4-x^2

    What x & y will give triangle max area and what is that area?

    I'm unsure of my first move to get an A(x).

    Edit: also to I need to split my triangle into halves for this?
    A = \frac{1}{2}bh

    A(x) = \frac{1}{2}(2x)(y) = x(4-x^2)
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  3. #3
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    great thanks I was afraid it was that simple.

    I found the A'(x) to get to the CV points.

    <br />
A'(x) = -3x^2+4<br />

    I know important parts of the parabola are max at (0,4) and y=0 at x=-2 & 2. So I know where the triangle contained. Do I plug the x values into A'(x)?
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  4. #4
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    I ended up with (-1.15, 2.68) and (1.15, 2.68) and an area of 3.08 units cubed. My graph of A'(x) and critical points with the original function all seem to fit. Is this correct?
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