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Math Help - Need help please have no clue

  1. #1
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    Need help please have no clue

    Suppose you are a pharmacist who manufactures pill capsules. The pill is shaped like a cylinder with two hemispheres on either end. The pill must have a volume of 1000mg. The surface area of the hemispheres cost twice as much as the surface area of the cylinder. Find the dimensions which will minimize the cost of the pill.


    Thanks for your help ahead of time
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  2. #2
    Super Member

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    Hello, Parrishguy!

    I'll get you started . . .


    Suppose you are a pharmacist who manufactures pill capsules.
    The pill is shaped like a cylinder with a hemisphere on both ends.
    The pill must have a volume of 1000 {\color{red}\rlap{///}}\text{mg} .
    maybe 1000 mm│ ?
    The surface area of the hemispheres cost twice as much as the surface area of the cylinder.
    Find the dimensions which will minimize the cost of the pill.
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          * - - - - + - - - - *
          |                   |
          |                   |
        h |                   | h
          |                   |
          |                   |
          * - - - - + - - - - *
                r       r
           *                 *
            *               *
              *           *
                  * * *
    The hemispheres and the cylinder have radius r.
    The cylinder has height h.

    The volume of the two hemispheres is: . \tfrac{4}{3}\pi r^3
    The volume of the cylinder is: . \pi r^2h

    The total volume is 1000: . \tfrac{4}{3}\pi r^3 + \pi r^2h \:=\:1000 \quad\Rightarrow\quad h \:=\:\frac{3000-4\pi r^3}{3\pi r^2} .[1]


    The surface area of the cylinder is: . 2\pi rh mm▓.
    If it costs k dollars per mm▓, its cost is: . 2\pi krh dollars.

    The surface area of the two hemispheres is: . 4\pi r^2 mm▓.
    Since it costs 2k dollars per mm▓. its cost is: . 8\pi kr^2 dollars.

    The total cost is: . C \;=\;8\pi kr^2 + 2\pi krh .[2]

    Substitute [1] into [2]: . C \;=\;8\pi kr^2 + 2\pi kr\left(\frac{3000-4\pi r^3}{3\pi r^2}\right)

    This simplifies to: . C \;=\;\tfrac{16}{3}\pi kr^2 + 2000kr^{-1}

    . . And that is the function we must minimize.

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