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Math Help - Volume of a box?

  1. #1
    Member realintegerz's Avatar
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    Volume of a box?

    Find the volume of a box as a function of the width of base of the box?

    If 12,000 cm^2 of material is available to make a box with a square base and an open top, find the volume of the box as a function of the width if the base of the box
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by realintegerz View Post
    Find the volume of a box as a function of the width of base of the box?

    If 12,000 cm^2 of material is available to make a box with a square base and an open top, find the volume of the box as a function of the width if the base of the box
    did you draw a diagram? (see the attached).

    let the width of the base be x and the height of the box be y

    thus the volume (length times width times height) is given by

    V = x^2 y

    Now we know the surface area is 12000, this is the sum of the areas of all the faces, so

    12000 = \underbrace{x^2}_{\text{area of base}} +  \underbrace{4xy}_{\text{area of the 4 sides}}

    that is 12000 = x^2 + 4xy

    now can you answer the problem?
    Attached Thumbnails Attached Thumbnails Volume of a box?-box7.gif  
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  3. #3
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    Hello, realintegerz!

    If 12,000 cm² of material is available to make a box
    with a square base and an open top, find the volume of the box
    as a function of the width of the base of the box.
    Code:
             *-------*
            /|      /|
           / |     / |
          *-------*  |
          |       |  |
          |       |  |
         h|       |  *
          |       | /
          |       |/ x
          *-------*
              x

    Let x = side of the square base.
    Let h = height of the box.

    The panels comprising the box is made up of:
    . . a base with area x^2
    . . and four sides with area xh each.
    The total surface area is: . x^2 + 4xh

    We are told that the total surface area will be 12,000 cm².

    So we have: . x^2 + 4xh \:=\:12,\!000 \quad\Rightarrow\quad h \:=\:\frac{12,\!000-x^2}{4x} .[1]


    The volume of the box is: . V \;=\;x^2h

    Substitute [1]: . V \;=\;x^2\left(\frac{12,\!000-x^2}{4x}\right) \quad\Rightarrow\quad \boxed{V \;=\;\frac{1}{4}(12,\!000x - x^3)}

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