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  1. #1
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    Calculus Help max

    A box with a rectangular bottom and no top is to be made from a rectangular piece of sheet metal with dimensions 16 by 30 inches created by cutting equal sized squares from the corners and then turning up the sides. What should be the dimensions of the squares if the bos is to have maximum volume?

    1. Description of what the diagram of the bos prior to cutting with the squares indicated in each corner identifying the unknown dimensions.

    2. Equation for the quantity being maximized.

    3. Dimensions of the square resulting in the maximum volume.

    Please help!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by frozenflames View Post
    A box with a rectangular bottom and no top is to be made from a rectangular piece of sheet metal with dimensions 16 by 30 inches created by cutting equal sized squares from the corners and then turning up the sides. What should be the dimensions of the squares if the bos is to have maximum volume?

    1. Description of what the diagram of the bos prior to cutting with the squares indicated in each corner identifying the unknown dimensions.
    See attachment

    2. Equation for the quantity being maximized.
    The volume of the box will be:

    V=a (16-2a) (30-2a)

    3. Dimensions of the square resulting in the maximum volume.
    This will be a solution of:

    dV/da=0.

    dV/da=d/da[4a^3 - 92a^2 + 480a]

    .........=12 a^2 - 184 a + 480

    So we want the roots of:

    12 a^2 - 184 a + 480=0,

    which are a=10/3, and a=12.

    The second of these roots is clearly non-physical, which leaves the first
    To check if this is a maximum we need to check that d^2V/da^2 is negative.

    d^2V/da^2 = 24 a - 184

    This is <0 when a=10/3.

    so we conclude that a=10/3 gives a maximum volume for the box.

    RonL
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