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Math Help - Largest rectangle that can inscribed in a semicircle?

  1. #1
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    Largest rectangle that can inscribed in a semicircle?


    Determine the area of the largest rectangle that can be inscribed in a semicircle of radius 8". Figure shows that the area can be written as A = (2x)y, if (x,y) is the point of the upper right corner of the rectangle. However, we choose to parameterize the area by a single value, the angle "theta".

    1)
    Derive the formula for the area of the inscribed rectangle as a function of theta. We refer to this function as A(theta) below.

    A(theta) = ?


    So I guess the area is: (2*x) * (sqrt(16-x^2))?


    2)
    Plot A(theta) over its relevant domain. What is the relevant domain?

    I have no idea what the domain is, is it [0, 8]?


    3)For what value of theta is the maximum area attained?


    4)What is the maximum area?



    I know I have to take the 1st derivative, but I don't know if my area is set up right. any help is greatly appreciated

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  2. #2
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    Quote Originally Posted by Lunar View Post
    Determine the area of the largest rectangle that can be inscribed in a semicircle of radius 8". Figure shows that the area can be written as A = (2x)y, if (x,y) is the point of the upper right corner of the rectangle. However, we choose to parameterize the area by a single value, the angle "theta".

    1) Derive the formula for the area of the inscribed rectangle as a function of theta. We refer to this function as A(theta) below.

    A(theta) = ?

    So I guess the area is: (2*x) * (sqrt(16-x^2))?


    2) Plot A(theta) over its relevant domain. What is the relevant domain?

    I have no idea what the domain is, is it [0, 8]?


    3)For what value of theta is the maximum area attained?


    4)What is the maximum area?



    I know I have to take the 1st derivative, but I don't know if my area is set up right. any help is greatly appreciated
    See the attached picture

    Largest rectangle that can inscribed in a semicircle?-capture.jpg

    The area of the whole rectangle will be 4 times the area of of the rectangle in the first quadrant.

    A(\theta)=4(8\cos(\theta)(8\sin(\theta)=256\sin(\t  heta)\cos(\theta)

    So from here just take the derivative.

    Also the the identity 2\sin(\theta)\cos(\theta)=\sin(2\theta)

    So A(\theta)=128\sin(2\theta)
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