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Math Help - Maximum possible area

  1. #1
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    Maximum possible area

    My first mistake was taking a Calculus class online... my second mistake was trying to learn Calculus from the textbook...


    Here is the problem that I'm having a hard time understanding:

    What is the maximum possible area of a rectangle inscribed into the ellipse
    [(x^2) + (4y^2) = 4] with the sides of the rectangle parallel to the coordinate axis?
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  2. #2
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    Quote Originally Posted by pennydooodle View Post
    My first mistake was taking a Calculus class online... my second mistake was trying to learn Calculus from the textbook...


    Here is the problem that I'm having a hard time understanding:

    What is the maximum possible area of a rectangle inscribed into the ellipse
    [(x^2) + (4y^2) = 4] with the sides of the rectangle parallel to the coordinate axis?
    Maximum possible area-capture.jpg

    from the diagram we see that the area of the rectangle in the first quadrant is A=xy so the total area is A=4xy

    solving for x in the ellipse we get x=2\sqrt{1-y^2}

    subbing into the Area formula we get

     A=8y \sqrt{1-y^2}

    taking the derivative

    \frac{dA}{dy}=8\left( \sqrt{1-y^2}+\frac{y}{2\sqrt{1-y^2}}(-2y)\right)=8\left( \frac{(1-y^2)-y^2}{\sqrt{1-y^2}}\right)

    setting the numerator equal to zero we get

    1-2y^2=0 \iff y=\frac{\sqrt{2}}{2}

    subbing y into the equation for x we get x=2\sqrt{1-\left( \frac{\sqrt{2}}{2}\right)^2}=2\sqrt{1-1/2}=\sqrt{2}

    So The area of the whole rectangle is A=4xy=4\sqrt{2}\frac{\sqrt{2}}{2}=4
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