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Math Help - Quickie #11

  1. #1
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    Quickie #11


    Across one corner of a rectangular room, two 4-foot screens
    are placed so as to enclose the maximum floor space.
    Determine their positions.
    Code:
          +----*-----
          |     *
          |      *
          |       *
          |        *
          |     *
          |  *
          *
          |
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Soroban View Post

    Across one corner of a rectangular room, two 4-foot screens
    are placed so as to enclose the maximum floor space.
    Determine their positions.
    Code:
          +----*-----
          |     *
          |      *
          |       *
          |        *
          |     *
          |  *
          *
          |
    Hello Soroban,

    unfortunately I can't offer you a complete solution..

    At first sight I would say that there must be 2 different solution to place those screens:
    a) they form with the walls a square with area = 16 sqft
    b) they form a straight line which is the base of an isosceles right triangle with area = 16 sqft

    At the attempt to get a more general solution I got lost completely:

    According to the attached sketch I got:

    A=\frac{x \cdot y}{2}+\frac{1}{2} \cdot \sqrt{x^2+y^2}\cdot \sqrt{4^2 - \left( \frac{1}{2}\sqrt{x^2+y^2} \right)^2}

    A=\frac{x \cdot y}{2}+\frac{1}{2} \cdot  \sqrt{16 (x^2+y^2) -  \frac{1}{4}(x^2+y^2)^2}

    A=\frac{x \cdot y}{2}+\frac{1}{4} \cdot  \sqrt{(x^2+y^2)\cdot (64-(x^2+y^2))}

    At the attempt to substitute y by a term in x I failed successfully. I only can ask you to ban my sorrow

    EB
    Attached Thumbnails Attached Thumbnails Quickie #11-ofenschirm.gif  
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  3. #3
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    Maybe a trick question?

    Each are 4 feet away from the corner to make a square of 16 square feet?

    Hehe...


    Edit: Or it could be infinity because when the area is enclosed, the area enclosed by the screens are on the outside??
    Last edited by anthmoo; January 14th 2007 at 02:09 AM.
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  4. #4
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    Nice try, EB . . . and a good guess, anthmoo!

    Remember this is a "Quickie",
    . . so there is usually a surprisingly simple solution.
    I found all of them to be valuable learning experiences.

    This problem has a head-slapping, OMG kind of solution . . . brace yourself.


    We have a pair of perpendicular walls and a pair of 4-foot screens.
    They could be arranged like this:
    Code:
          * - - -*- - - - - -
          |       *
          |        *
          |         *
          *          *
          |   *        *
          |       *     *
          |           *  *
          |               *

    Now consider four of these figures clustered about the right angle.
    Code:
          *       |       *
           *  *   |   *  *
            *     *     *
             *    |    *
          - - * - + - * - - 
             *    |    *
            *     *     *
           *  *   |   *  *
          *       |       *

    We have an octagon with eight equal sides.

    Obviously an octagon with a fixed perimeter has maximum area
    . . when it is a regular octagon.


    Therefore, the screens are placed like this:
    Code:
          *-----------------*--
          |                *
          |               *
          |              *
          |             *
          |      135  *
          |           *
          |       *
          |   *
          *
          |
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