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Math Help - Optimizing Problem of a Window

  1. #1
    Newbie RangerKimmy's Avatar
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    Optimizing Problem of a Window

    "A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 22 ft, find the dimensions of the window so that the greatest possible amount of light is admitted."

    (I've seen some forum posts about optimizing Norman Windows, but I've still gotten stuck. I was hoping to get some help with my problem.)

    EDIT: Okay, I'm still having trouble, even though I've reworked it to fit the problem properly and solve for optimized dimensions.

    Perimeter:  \pi r + 2h + 2r \:=\:22\quad\Rightarrow\quad h \:=\:\frac{22 - 2r - \pi r}{2}

    Substituting:  P \;=\;\pi r + 2\left(\frac{22 - 2r - \pi r}{2}\right) + 2r

    And now I have trouble. I know I needed to plug in my solved h to eliminate a variable, then simplify, and then take the derivative.
    But simplifying it brings it back to P = 22, which... isn't what I need.

    Simplifying:  P \;=\;\pi r + 22 - 2r - \pi r + 2r

     P \;=\; 22

    Clearly this isn't the right method! I don't know what I'm doing wrong. Usually I'm supposed to work with Area too, but that shouldn't be involved this time, should it?
    Last edited by RangerKimmy; November 27th 2009 at 07:00 PM. Reason: Read the question incorrectly. Fixed now.
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  2. #2
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    You're right

    The calculation looks good. Don't be scared of
    <br />
\pi + 4<br />
    It's just the exact value of a number equal to about 7.14159...
    Your answer will be as precise as the approximation you use. For a rough estimate, just use
    <br />
\pi + 4 \approx 7<br />
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  3. #3
    Newbie RangerKimmy's Avatar
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    Quote Originally Posted by qmech View Post
    The calculation looks good. Don't be scared of
    <br />
\pi + 4<br />
    It's just the exact value of a number equal to about 7.14159...
    Your answer will be as precise as the approximation you use. For a rough estimate, just use
    <br />
\pi + 4 \approx 7<br />
    Thanks, but it turns out I read the question incorrectly.. I was looking for dimensions, not area. And now I have entirely NO idea how to do this problem because I am used to working with area, not dimensions.
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  4. #4
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    mr fantastic's Avatar
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    Quote Originally Posted by RangerKimmy View Post
    Thanks, but it turns out I read the question incorrectly.. I was looking for dimensions, not area. And now I have entirely NO idea how to do this problem because I am used to working with area, not dimensions.
    Go back to your calculations. Have you found the height and width of the rectangle? These give you the dimensions.
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  5. #5
    Newbie RangerKimmy's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Go back to your calculations. Have you found the height and width of the rectangle? These give you the dimensions.
    I'm still doing this incorrectly..
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