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Math Help - Finding the minimum cost for a semi-circular cover

  1. #1
    Junior Member Cthul's Avatar
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    Finding the minimum cost for a semi-circular cover


    Not sure how to do the last bit, help please.
    Note: The cover is sometimes referred as a lid for some reason.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You didn't show the right equation for part (c)... =/

    Area of two end sections = \pi r^2

    Area of body of cover = \dfrac12 2\pi r h = \pi rh[/tex]

    Cost of end sections = 8\pi r^2

    Cost of body = 4\pi rh

    Total cost = 8\pi r^2 + 4 \pi rh

    Now,  h = \dfrac{16}{\pi r^2}

    Hence, cost = 8\pi r^2 + 4 \pi r\left( \dfrac{16}{\pi r^2}\right) = 8\pi r^2 + \left( \dfrac{64}{ r}\right)

    4. I hope that you know some calculus... you'll have to find \dfrac{dC}{dr} = 0 then solve for r.
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  3. #3
    Junior Member Cthul's Avatar
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    I know how to do calculus. Thanks.
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