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Math Help - Optimization- Word Problem, minimum amount of material to constuct

  1. #1
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    Optimization- Word Problem, minimum amount of material to constuct

    A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a capacity of , find the radius and height of the silo that requires the least amount of material to construct.
    Hint: The volume of the silo is , and the surface area (including the floor) is

    ... I need to find "r" and "h"
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  2. #2
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    Hello, franklen!

    A grain silo has the shape of a right circular cylinder surmounted by a hemisphere.
    If the silo is to have a capacity of 614\pi ft³, find the radius rand height h of the silo
    that requires the least amount of material to construct.

    Hint: The volume of the silo is: . V \:=\;\pi r^2h + \frac{2}{3}\pi r^3 .[1]
    . . . and the surface area (including the floor) is: . S \:=\:3\pi r^2 + 2\pi rh .[2]
    They gave us the necessary formulas . . .


    From [1], we have: . V \:=\:\pi r^2h + \frac{2}{3}\pi r^3 \:=\:614\pi \quad\Rightarrow\quad h \:=\:614r^{-2} + \frac{2}{3}r .[3]

    Substitute into [2]: . S \;=\;3\pi r^2 + 2\pi r\left(614r^{-2} + \frac{2}{3}r\right)
    . . which simplifies to: . S \;=\;\frac{13}{3}\pi r^2 + 1228\pi r^{-1}

    Differentiate and equate to zero: . S' \;=\;\frac{26}{3}\pi r - 1228\pi r^{-2} \;=\;0

    Muliply by 3r^2\!:\;\;26\pi r^3 - 3684\pi \;=\;0

    Then we have: . r^3 \;=\;\frac{3684\pi}{26\pi} \;=\;\frac{1842}{13} \quad\Rightarrow\quad \boxed{r \;=\;\sqrt[3]{\frac{1842}{13}}}


    From .[3], we have: . h \;=\;\frac{1842 + 2r^3}{3r^2}

    Substitute: . h \;=\;\frac{1842 + 2\left(\frac{1842}{13}\right)}{3\left(\frac{1842}{  13}\right)^{\frac{2}{3}}}

    . . and after a LOT of algebra, we get: . \boxed{h \;=\;5\sqrt[3]{\frac{1842}{13}}}


    Fascinating . . . the height is exactly five times the radius.
    .
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  3. #3
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    hmm... it says that those answers are incorrect?
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