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Math Help - Optimization [Derivatives]

  1. #1
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    Optimization [Derivatives]

    What is the largest rectangle that can be inscribed in the first quadrant of the ellipse

    I let x and y equal 0 to find the intercepts, but that does not lead to the solution.
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  2. #2
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    Quote Originally Posted by castle View Post
    What is the largest rectangle that can be inscribed in the first quadrant of the ellipse

    I let x and y equal 0 to find the intercepts, but that does not lead to the solution.
    1. Draw a sketch.

    2. Let l denote the length of the rectangle and w the width. Then the area of the rectangle is:

    a = l \cdot w

    3. Plug in l instead of x and calculate the y-value which is the width of the rectangle:

    w = \frac34 \sqrt{16-l^2}

    4. You now got the area as a function in l:

    a(l)= l \cdot \frac34 \sqrt{16-l^2} = \frac34 \cdot l \cdot \sqrt{16-l^2}

    5. Differentiate a wrt l and solve the equation a'(l) = 0 for l. I've got l = 2 \sqrt{2}

    6. Now calculate the maxium area.
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    You can always go with Lagrange multipliers. The equation of constraint is g=9x^2+16y^2-144

    The area is A=xy

    You have \nabla A=\lambda\nabla g

    \left(y,x\right)=\lambda\left(18x,32y\right)

    y=18\lambda x

    x=32\lambda y=32\lambda\times 18\lambda x=24^2\lambda^2x

    \lambda=\pm\dfrac{1}{24}

    y=\pm\dfrac{3}{4}x

    g=9x^2+16\dfrac{9}{16}x^2-144=0

    18x^2=144

    x=\pm \sqrt{8} but x>0 so x=\sqrt{8}

    y=\pm \dfrac{3}{4}x=\pm\dfrac{3}{4}\times \sqrt{8}=\dfrac{3}{\sqrt{2}}
    Last edited by fobos3; July 14th 2010 at 11:31 AM.
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  4. #4
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    Quote Originally Posted by earboth View Post

    w = \frac34 \sqrt{16-l^2}

    I know it is a silly question, but can you show me step by step how you got that. Thank You.
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  5. #5
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    Quote Originally Posted by castle View Post
    I know it is a silly question, but can you show me step by step how you got that. Thank You.
    Here we go:

    Solve for y:

    \begin{array}{rcl}9l^2+16w^2&=&144 \\ 16w^2&=&144-9l^2 \\ w^2&=&\dfrac{144-9l^2}{16} \\ w^2&=&\dfrac{9(16-l^2)}{16}\end{array}

    ... and now calculate the square-root of both sides of the equation.
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  6. #6
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    O.K. Thanks.
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