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Math Help - Word Problem with a Graph

  1. #1
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    Word Problem with a Graph

    1) What is the area of the largest rectangle below the graph of the function f(x)=4-3x^2 and above the x-axis with its base on the x-axis?
    -I am unable to post the graph, but I am able to find the equation of the area to be 2z(4-3z^2). However, I do not know how to continue the problem.

    Any help will be much appreciated!

    -M
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  2. #2
    Eater of Worlds
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    y=4-3x^{2}

    The area of the rectangle is A=2xy

    Therefore, A=2x(4-3x^{2})=8x-6x^{3}

    A'(x)=8-18x^{2}

    8-18x^{2}=0

    x=\pm{\frac{2}{3}}

    y=\pm{\frac{8}{3}}
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  3. #3
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    Quote Originally Posted by blurain View Post
    1) What is the area of the largest rectangle below the graph of the function f(x)=4-3x^2 and above the x-axis with its base on the x-axis?
    -I am unable to post the graph, but I am able to find the equation of the area to be 2z(4-3z^2). However, I do not know how to continue the problem.

    Any help will be much appreciated!

    -M
    f(x) = 4 -3x^2 is "vertical" parabola that opens dowwnward, whose vertex is (0,4), whose axis of symmetry is x=0 or the f(x)=y axis.

    So, area A = 2x(y)
    A = 2x(4 -3x^2) -------------not z's.

    The largest A is when dA/dx = 0, so,
    dA/dx = 2x[-6x] +(4 -3x^2)[2]
    0 = -12x^2 +8 -6x^2
    0 = -18x^2 +8
    x^2 = 8/18 = 4/9
    x = 2/3

    So, max A = 2(2/3)[4 - 3(2/3)^2]
    max A = (4/3)[4 - 4/3]
    max A = (4/3)[(3*4 -4)/3]
    max A = (4/3)[8/3]
    max A = 32/9 sq.units ------------answer.
    In decimals,
    max A = 3.555....sq.units
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