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Math Help - 2 tricky optimization problems

  1. #1
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    Question 2 tricky optimization problems

    Hey guys,
    I'm stuck with these two problems, I tried various ways of solving, but with no luck.

    1. A rectangular page is to contain 180 square inches of print. The top and bottom margins are each 0.7 inches wide , and the margins on each side is 2 inches wide. What should the dimensions be if the least amount of material is to be used ?

    Length of top(bottom) :
    Length of side :

    2. Suppose postal requirements are that the maximum of the length plus the girth (cross sectional perimeter) of a rectangular package that may be sent is 250 inches. Find the dimensions of the package with square ends whose volume is to be maximum.

    Square side :
    Length :

    What to do? Thanks.

    PS: solutions are due 8 pm today june 12th!
    Last edited by Starmix; June 12th 2009 at 01:40 PM.
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  2. #2
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    Hello, Starmix!

    Who made up these problems?
    The answers are quite ugly.


    1. A rectangular page is to contain 180 inē of print.
    The top and bottom margins are each 0.7 in wide, and the margins on each side is 2 in wide.
    What should the dimensions be if the least amount of material is to be used?
    Did you make a sketch?
    Code:
          : 2 : - x - : 2 :
        - *---------------* -
      0.7 |               | :
        - |   *-------*   | :
        : |   |       |   | :
        : |   |       |   | :
        y |   |y      |   | : y + 1.4
        : |   |       |   | :
        : |   |   x   |   | :
        - |   *-------*   | :
      0.7 |               | :
        - *---------------* -
          : - - x + 4 - - :

    The width of the print is x.
    The height of the print is y.
    The area of the print is 180 inē: . xy = 180 \quad\Rightarrow\quad y = \tfrac{180}{x} .[1]

    The width of the page is: x+4.
    The height of the page is: y + 1.4.
    The area of the page is: . A \;=\;(x+4)(y+1.4) .[2]


    Substitute [1] into [2]: . A \;=\;(x + 4)\left(\tfrac{180}{x} + 1.4\right)

    . . which simplifies to: . A \;=\;1.4x + 720x^{-1} + 185.6


    Then we have: . A' \;=\;1.4 - 720x^{-2} \:=\:0 \quad\Rightarrow\quad 1.4x^2 - 720 \:=\:0

    . . Then: . x^2 \:=\:\tfrac{720}{1.4} \:=\:\tfrac{3600}{7} \quad\Rightarrow\quad x \:=\:\tfrac{60}{\sqrt{7}} \:=\:\frac{60\sqrt{7}}{7}

    Substitute into [1]: . y \:=\:\frac{180}{\frac{60}{\sqrt{7}}} \:=\:3\sqrt{7}


    The width of the page is: . x + 4 \:=\:\frac{60\sqrt{7}}{7} + 4

    The height of the page is: . y + 1.4 \:=\:3\sqrt{7} + 1.4

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  3. #3
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    Merci.....! )

    Actually I did everything the same except for the margins...I subtracted them from x and y values instead of adding.
    Last edited by mr fantastic; June 13th 2009 at 01:48 AM. Reason: Merged posts
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  4. #4
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    Hello again, Starmix!

    Actually I did everything the same except for the margins.
    I subtracted them from x and y values instead of adding.

    I did, too . . . but the algebra/arithmetic got quite messy.

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  5. #5
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    Hello, Starmix

    A bit late, but here's the other one . . .


    2. Suppose postal requirements are that the maximum of the length plus the girth
    (cross sectional perimeter) of a rectangular package that may be sent is 250 inches.
    Find the dimensions of the package with square ends whose volume is to be maximum.
    Code:
                * - - - *
               /       /|
              /       / |
             /       /  |
            /       /   *
           /       /   /
          * - - - *   /
          |       |  / y
        x |       | /
          |       |/
          * - - - *
              x

    The girth is 4x, the length is y.
    . . 4x + y \:=\:250 \quad\Rightarrow\quad y \:=\:250-4x .[1]

    The volume of the box is: . V \;=\;x^2y .[2]


    Substitute [1] into [2]: . V \;=\;x^2(250-4x) \;=\;250x^2 - 4x^3

    Then: . V' \;=\;500x - 12x^2 \:=\:0 \quad\Rightarrow\quad 4x(125 - 3x) \:=\:0

    And we have: . \begin{Bmatrix}x \:=\:0 & \text{min.volime} \\ x \:=\:\frac{125}{3} & \text{max.volume} & {\color{red}*}\end{Bmatrix}

    Substitute into [1]: . y \:=\:250 - 4\left(\tfrac{125}{3}\right) \:=\:\tfrac{250}{3}


    Square side: . 41\tfrac{2}{3} inches.

    . . . .Length: . 83\tfrac{1}{3} inches.

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  6. #6
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    Quote Originally Posted by Starmix View Post
    Hey guys,
    I'm stuck with these two problems, I tried various ways of solving, but with no luck.

    1. A rectangular page is to contain 180 square inches of print. The top and bottom margins are each 0.7 inches wide , and the margins on each side is 2 inches wide. What should the dimensions be if the least amount of material is to be used ?

    Length of top(bottom) :
    Length of side :

    2. Suppose postal requirements are that the maximum of the length plus the girth (cross sectional perimeter) of a rectangular package that may be sent is 250 inches. Find the dimensions of the package with square ends whose volume is to be maximum.

    Square side :
    Length :

    What to do? Thanks.

    PS: solutions are due 8 pm today june 12th!
    (My highlighting) I assume your teacher has no problem with you getting people to do your homework for you.
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