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Math Help - minimization of a water tank

  1. #1
    Junior Member winterwyrm's Avatar
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    minimization of a water tank

    Thanks in advance, it's attached. In case you can't read it, the problem states: The tank below is to be constructed to hold 5.00 x 10^5L when filled completely. The shape is cylindrical with a hemispherical top. The cost is a function of surface area, it will cost $300/(m^2) for the cylindrical portion, and $400/(m^2) for the hemispherical portion. Ignore the cost to construct the bottom circular area.

    Specify the dimensions, H & R which will result in the lowest dollar cost, where H is the height of the cylinder, and R is the radius of the sphere and of the cylinder.
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    The volume of the tank is:

    V=\frac{2}{3}{\pi}r^{3}+{\pi}r^{2}h=500,000......[1]

    The surface area of the hemisphere is given by 2{\pi}r^{2}

    The surface area of the cylidrical portion, excluding the base is given by:

    2{\pi}rh

    So, the total surface area is:

    S=2{\pi}r^{2}+2{\pi}rh

    But, the cost for the hemsphere is $400 and the cylinder is $300:

    S=400(2{\pi}r^{2})+300(2{\pi}rh)....[2]

    Solve [1] for, say, h and sub into [2]. Then differentiate, set to 0 and solve for r. h will follow.
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  3. #3
    Junior Member winterwyrm's Avatar
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    when I do it this way though, I get R= 457.08m and H= 913.08m though, and those just don't work in the original equation, is this right?
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