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Math Help - Optimisation of area

  1. #1
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    Optimisation of area

    Hi guys, heres another great optimisation problem Just having a bit of trouble getting the right equations etc any help appreciated. Also as is stated in the brackets a net of the shape is required. As usual any help is appreciated. Thanx.

    Question: Consider a rectangle of perimeter 60Cm. Form a cylinder by revolving this rectangle about one of its edges. What dimensions of the rectangle will result in a cylinder of maximum volume? (Include a net of the final shape)
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  2. #2
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    To find: "What dimensions of the rectangle will result in a cylinder of maximum volume?"
    So the dimensions of the rectangle when the first derivative of the volume of the cylinder is zero.

    There are two possibilities: one is when the rectangle is revolved about one of its horizontal edges, and the other is when the rectangle is revolved about one of its vertical edges---if the 4 edges were either horizontal or vertical.

    --------------------
    1) When revolved about a horizontal edge.

    Let vertical edges be x cm each, so the horizontal edges are (30-x) cm each.

    Volume of generated cylinder, V1 = "pi(r^2)L", where r=x, and L = (30-x) here.
    So,
    V1= pi(x^2)(30-x)
    V1 = pi(30x^2 -x^3)
    Differentiate both sides with respect to x,
    dV1/dx = pi[60x -3x^2]
    Set that to zero,
    0 = 60x -3x^2
    Divide both sides by 3x,
    0 = 20 -x
    x = 20 cm ----------vertical side or edge.
    (30-x) = 10 cm ----horizontal edge.

    Therefore, the rectangle must be 20cm by 10cm. -------answer.
    (max V1 = pi(20^2)(10) = 4000pi cu.cm.)

    The net of the generated cylinder?
    The surface area of the cylinder is a rectangle that is (10cm wide) by (2pi*20 cm long), or a rectangle 20cm by 40pi cm.

    ----If the net in question refers only to this suface area , then the net is a rectangle 10cm by 40pi cm.
    ----If the net in question includes also the 20cm by 10cm original rectangle,then the net is composed of two rectangles, side by side at their respective 10cm sides, a 10-by-20 and a 10-by-40pi. Or it is a longer rectangle that is 10cm by (20 +40pi)cm.
    ----If the net in question includes also the circular planes at each vertical side of the cylinder, then include two circles of radius 20cm each.

    ------------------------------------------------------------
    1) When revolved about a vertical edge.

    Let again vertical edges be x cm each, so the horizontal edges are (30-x) cm each.

    r = (30-x), and L=x here.
    So, V2 = pi[(30-x)^2](x)
    V2 = pi[900 -60x +x^2](x)
    V2 = pi[900x -60x^2 +x^3]
    Differentiate both sides with respect to x,
    dV2/dx = pi[900 -120x +3x^2]
    Set that to zero,
    0 = 900 -120x +3x^2
    Divide both sides by 3,
    x^2 -40x +300 = 0
    Factor that,
    (x-10)(x-30) = 0
    x = 10 or 30

    When x=30, the horizontal edge (30-x) becomes zero, so no good.

    Therefore, x = 10 cm ------vertical edge of the rectangle.
    (30-x) = 20 cm -----------horizontal edge rectangle.

    Therefore, the rectangle must be 10cm by 20cm. -----answer.
    (max V2 = pi(20^2)(10) = 4000pi cu.cm. also, same as max V1)

    The net of the generated cylinder?
    The same as in V1.
    Last edited by ticbol; August 20th 2007 at 01:59 AM. Reason: volume is in cu.cm., not in sq.cm.
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  3. #3
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    Thanx for that mate
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  4. #4
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    Hello, mathsB!

    I have no idea what a "net" is . . .


    Consider a rectangle of perimeter 60Cm.
    Form a cylinder by revolving this rectangle about one of its edges.
    What dimensions of the rectangle will result in a cylinder of maximum volume?
    (Include a net of the final shape)
    We have a rectangle with dimensions L \times W.
    Code:
          * - - - - - *
          |           |
        W |           |
          |           |
          * - - - - - *
                L
    The perimeter is 60 cm: . 2L + 2W \:=\:60\quad\Rightarrow\quad W \:=\:30 - L .[1]

    Revolve the rectangle about a vertical side.
    The cylinder has radius L and height W.
    . . It volume is: . V \:=\:\pi r^2h \:=\:\pi L^2W. . [2]

    Substitute [1] into [2]: . V \;=\;\pi L^2(30-L)


    Now maximize this function.
    . . \text{(I got: }L = 20,\;W = 10)

    Last edited by Soroban; August 19th 2007 at 06:27 AM.
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  5. #5
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    Quote Originally Posted by Soroban View Post

    I have no idea what a "net" is . . .


    ...
    Hello,

    only for the record: I've attached a "net" of a cube and a "net" of perpendicular circular cylinder. Maybe you now can provide me with the correct English expression
    Attached Thumbnails Attached Thumbnails Optimisation of area-zyl_cube_netz.gif  
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