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Math Help - Calculus Maximize Help

  1. #1
    Bar0n janvdl's Avatar
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    Exclamation Calculus Maximize Help

    Hi guys i hav an urgent prob.

    I have to hand in a container of 1 litre on monday, but im not sure if my calculations r correct.

    The container shud have a length twice the width.
    It shud have a vol of 1000 ml
    It shud hav a minimum exterior area.

    Ok this is what i did.

    L = 2x
    B = x
    H = x
    Vol = L . B . H

    1000 ml = 2x^3

    We dont have to find the derivative if we hav a max value for the volume already or do we?


    K, the exterior area:
    Ex-Area = (2L+B)H + 2(LxB)
    In the end the Ex Area = 10x^2

    But what now? Shud we find the derivative?
    Pls help me.
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  2. #2
    Bar0n janvdl's Avatar
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    Oh yeah i found the B and H = 7,93700526
    L = 2 x 7,93700526
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by janvdl View Post
    Hi guys i hav an urgent prob.

    I have to hand in a container of 1 litre on monday, but im not sure if my calculations r correct.

    The container shud have a length twice the width.
    It shud have a vol of 1000 ml
    It shud hav a minimum exterior area.

    Ok this is what i did.

    L = 2x
    B = x
    H = x
    Vol = L . B . H

    1000 ml = 2x^3
    L=2x, B=x, H=h, so volume V= 2x^2 h =1000

    We dont have to find the derivative if we hav a max value for the volume already or do we?


    K, the exterior area:
    Ex-Area = (2L+B)H + 2(LxB)
    K= 2(L.H) + 2(B.H) + 2(L.B) = 4x.h + 2x.h + 2x^2 = 6x.h + 2x^2

    but from the known volume we have x.h = 1000/x, so:

    K = 6000/x + 2x^2

    To maximize K we look for solutions to dK/dx = 0.

    dK/dx = -6000/x^2 + 4x

    so we want solutions of -6000/x^2 + 4x=0, or 6000=4x^3, so

    x=cuberoot(1500) ~= 11.45 cm

    RonL
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  4. #4
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    Hello, janvdl!

    I suggest you use more familiar variables . . .


    The rectangular container should have a length twice the width.
    It should have a vol of 1000 ml. .It should have a minimum exterior area.

    Let L = length, W = width, H = height of the container.

    Since the length is twice the width: .L = 2W

    The box looks like this:
    Code:
                * - - - - - - - *
              /               / |
            /               /   | H
          * - - - - - - - *     |
          |               |     *
        H |               |   /W
          |               | /
          * - - - - - - - *
                 2W

    The volume of a "box" is: .V .= .L W H

    So we have: .V .= .(2W)(W)(H) .= .1000 . . H .= .500/W .[1]


    Now determine the exterior area of the box.

    . . Front and back panels: .2 (2W)(H) .= .4WH

    . . Left and right panels: .2 (W)(H) .= .2WH

    . . Top and bottom panels: .2 (2W)(W) .= .4W

    Hence, the total exterior area is: .A .= .4W + 6WH .[2]


    Substitute [1] into [2]: .A .= .4W + 6W(500/W)

    . . and we have: .A .= .4W + 3000W^
    {-1}


    And that is the function we must minimize . . .

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