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Math Help - Finding the Maximum Area of a cylinder

  1. #1
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    Finding the Maximum Area of a cylinder

    Here's the problem. Can someone work through it for me? I don't know where to start.

    A cylinder is inscribed in a right circular cone of height 4 and radius (at the base) equal to 4.5. What are the dimensions of such a cylinder which has maximum volume?
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  2. #2
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    Hello, lauren72!

    Did you make a sketch?


    A cylinder is inscribed in a right circular cone of height 4 and base radius 4.5.
    What are the dimensions of such a cylinder which has maximum volume?
    Look at a cross-section of the cone.
    Code:
        -                 *
        :               * | *
        :             *   |   *
        :           *     |     *
        4         * - - - + - - - *
        :       * |       |       | *
        :     *   |       |h      |   *
        :   *     |       |       |     *
        - *-------+-------+-------+-------*
          :      4.5      :   r   : 4.5-r :

    The radius of the cylinder is r, its height is h.

    From the similar right triangles: . \frac{h}{\frac{9}{2}-r} \:=\:\frac{4}{\frac{9}{2}} \quad\Rightarrow\quad h \:=\:\frac{4}{9}(9-2r) .[1]

    The volume of a cylinder is: . V \;=\;\pi r^2 h .[2]

    Substitute [1] into [2]: . V \;=\;\pi r^2\cdot\frac{4}{9}(9-2r) \quad\Rightarrow\quad V \;=\;\frac{4\pi}{9}(9r^2-2r^3)

    Differentiate and equate to zero: . V' \;=\;\frac{4\pi}{9}(18r - 6r^2) \;=\;0

    Then: . 18r-6r^3 \:=\:0 \quad\Rightarrow\quad 6r(3-r) \:=\:0

    Hence: . \begin{Bmatrix}r \:=\:0 & \text{minimum volume} \\ r \:=\:3 & \text{maximum volume} \end{Bmatrix}\quad\Rightarrow\quad r \,=\,3

    Substitute into [1]: . h \:=\:\frac{4}{9}(9-2\!\cdot\!3) \:=\:\frac{4}{3}


    Therefore: . r = 3,\;h = \frac{4}{3}

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  3. #3
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    You, my friend, are a saint.
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