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Math Help - Box Dimensions

  1. #1
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    Box Dimensions

    The dimensions of a gift box are consecutive positive integers such that the height is the least integer and the length is the greatest integer. If the height is increased by 1cm, the width is increased by 2cm, and the length is increased by 3cm, then a larger box is constructed such that the volume is increased by 456cm3. determine the dimensions of each box.

    and so far I got to
    LWH=V
    (L+3)(W+2)(H+1)= V+456

    .___. then I got stuck >< I don't even know if I'm on the right track.
    According to the answers in the back, the dimensions of the first box are
    9cm by 8cm by 7cm
    the second box would be
    12cm by 10cm by 8cm

    Thank you in advance for your help ^^
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  2. #2
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    Quote Originally Posted by orangepower View Post
    The dimensions of a gift box are consecutive positive integers such that the height is the least integer and the length is the greatest integer.
    Therefore w = h+1 and l = h+2
    If the height is increased by 1cm, the width is increased by 2cm, and the length is increased by 3cm, then a larger box is constructed such that the volume is increased by 456cm3. determine the dimensions of each box.

    and so far I got to
    LWH= h(h+1)(h+2) = h^3+3h^2+2h = V
    ((h+2)+3)((h+1)+2)(H+1)= V+456

    .___. then I got stuck >< I don't even know if I'm on the right track.
    According to the answers in the back, the dimensions of the first box are
    9cm by 8cm by 7cm
    the second box would be
    12cm by 10cm by 8cm

    Thank you in advance for your help ^^
    I've added some necessary information
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  3. #3
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    Hello, orangepower!

    You overlooked the most powerful condition in the problem.


    The dimensions of a gift box are consecutive positive integers
    such that the height is the least integer and the length is the greatest integer.
    If the height is increased by 1 cm, the width is increased by 2 cm,
    and the length is increased by 3 cm,
    then a larger box is constructed such that the volume is increased by 456 cm³.
    Determine the dimensions of the box.

    We have: . \begin{array}{ccc}\text{Height} &=& n \\ \text{Width} &=& n+1 \\ \text{Length} &=& n+2 \end{array}

    The original volume is: . V_o \:=\:n(n+1)(n+2) \:=\:n^3 + 3n^2 + 2n


    The larger box has dimensions: . \begin{array}{ccc}\text{Height} &=& n+1 \\ \text{Width} &=& n+3 \\ \text{Length} &=& n+5\end{array}

    The larger volume is: . V_1 \:=\:(n+1)(n+3)(n+5) \:=\:n^3 + 9n^2 + 23n + 15


    The larger volume is 456 cm³ more than the original volume.
    . . n^3 + 9n^2 + 23n + 15 \;=\;n^3 + 3n^2 + 2n + 456

    This simplifies to: . 6n^2 + 21n - 441 \:=\:0 \quad\Rightarrow\quad 2n^2 + 7n - 147 \:=\:0

    . . which factors: . (n-7)(2n+21) \:=\:0

    . . and has roots: . n \:=\:7,\:\text{-}\tfrac{21}{2}


    Therefore, the dimensions of the box are: . 7\text{ by }8\text{ by }9

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  4. #4
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    ^ I know how you found 7 but how did you find 8 and 9? D:

    ahhh >< I finally got it THANKS!!!
    Last edited by orangepower; March 10th 2009 at 03:00 AM.
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