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Math Help - Box Construction

  1. #1
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    Box Construction

    (1) An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?

    (2) Using the question above, solve if the piece of sheet metal is a rectangle whose length is twice the width.
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  2. #2
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    Quote Originally Posted by symmetry View Post
    (1) An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?

    (2) Using the question above, solve if the piece of sheet metal is a rectangle whose length is twice the width.
    (1)The height of the box will be 1 ft, so the area of the base must be 4 sq ft
    So if the width of the box is w ft, the length is 4/w ft, and the size of the
    sheet it is floded from is w+2 by (4/w)+2 ft.

    Opps, did not notice that we has a square sheet of metal, as its square w=2, and the
    original sheet is as Earboth says 4'x4'

    (2)Here 2(w+2)=(4/w)+2, which you should solve for w, you will have two
    roots one of which will not satisfy the conditions to be a width, and the
    other will, and it will be the solution.

    RonL
    Last edited by CaptainBlack; January 13th 2007 at 11:24 AM.
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  3. #3
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    Quote Originally Posted by symmetry View Post
    (1) An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?

    (2) Using the question above, solve if the piece of sheet metal is a rectangle whose length is twice the width.
    Hello,

    to (1):

    the volume is V=A_{base} \cdot h

    the height is 1'. thus the base area must be 4 sqft. That means the base area has the dimension 2'\times 2'. Therefore the piece of metal must have the dimension 4'\times 4'

    to(2):

    let the width of the piece be x
    then the length of the piece is 2x
    If you cut off squares of 1'\times 1' the base area must have the same value:

    (x-2)(2x-2)=4. Expand the LHS and solve for x. I've got x = 0 or x = 3.

    That means the piece of metal has the dimension 3'\times 6'.

    EB
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