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Math Help - Area of a rectangle within an ellipse

  1. #1
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    Area of a rectangle within an ellipse

    Find the dimensions of the largest rectangle wish dies parallel to the axes that can be inscribed in the ellipse x^2 + 4 y^2 = 4.

    Okay, here's my work so far:
    A=4xy
    Rearrange for x so A=4y(sqrt(4-4y^2))
    Use product rule, obtain derivative and set A' = 0
    It eventually comes to A' = -32y^2/2 (sqrt (4-4y^2)) + 4 (sqrt (4-4y^2))
    And I solve to get y = sqrt2/2, which is incorrect.
    The answer for the dimensions is 2sqrt2 by sqrt2. I know I have to sub to get x back after, but is there some algebraic process which is incorrect here? Any help would be super.
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  2. #2
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    Hello, cinematic!

    Your work is correct . . . up to the punchline.
    . . Did you forget to "double"?


    Find the dimensions of the largest rectangle with sides parallel to the axes
    that can be inscribed in the ellipse: x^2 + 4 y^2 \:=\: 4
    Code:
                        | 
                   *    *    *
              *         |         *
            o - - - - - + - - - - - o
           *:           |           :*
            :           |          y: 
          * :           |           : *
      - - * : - - - - - + - - - - - : * - -
          * :           |     x     : *
            :           |           :
           *:           |           :* 
            o - - - - - + - - - - - o
              *         |         *
                   *    *    *
                        |

    The upper half of the ellipse is: . y \:=\:\sqrt{1-\frac{x^2}{4}}

    The area of the rectangle is: . A \;=\;4xy

    We have: . A \;=\;4x\left(1 - \tfrac{1}{4}x^2\right)^{\frac{1}{2}}


    Then: . A' \;=\;4x\cdot\tfrac{1}{2}\left(1 - \tfrac{1}{4}x^2\right)^{\text{-}\frac{1}{2}}\! \left(\text{-}\tfrac{1}{2}x\right) + 4\left(1-\tfrac{1}{4}x^2\right)^{\frac{1}{2}}

    . . . -\frac{x^2}{\sqrt{1-\frac{x^2}{4}}} + 4\sqrt{1 - \tfrac{x^2}{4}} \;=\;0 \quad\Rightarrow\quad \frac{x^2}{\sqrt{1-\frac{x^2}{4}}} \;=\;4\sqrt{1-\tfrac{x^2}{4}}

    . . . x^2 \:=\:4\left(1 - \tfrac{x^2}{4}\right) \quad\Rightarrow\quad x^2 \:=\:4-x^2 \quad\Rightarrow\quad 2x^2 \:=\:4 \quad\Rightarrow\quad x^2 \:=\:2

    Hence: . x \:=\:\sqrt{2} \quad\Rightarrow\quad y \:=\:\frac{\sqrt{2}}{2}


    Therefore: . \begin{Bmatrix}\text{width} &=& 2x &=& 2\sqrt{2} \\ \text{height} &=& 2y &=& \sqrt{2} \end{Bmatrix}

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