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Math Help - Using the Derivative to Analyze Polynomial Function Model Problems

  1. #1
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    Using the Derivative to Analyze Polynomial Function Model Problems

    Can someone please help me with this question???

    Determine the dimensions of the cone of maximum volume that can be inscribed in a sphere with radius 3?
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  2. #2
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    745
    Hello, philipsach1!

    Did you make a sketch?


    [I] Determine the dimensions of the cone of maximum volume
    that can be inscribed in a sphere with radius 3?
    Code:
                    A
                  * * *
              *    /:\    *
            *     / : \     *
           *     /  :  \     *
                /   :3  \
          *    /    :    \    *
          *   /    O*     \   *
          *  /  3*  :  *3  \  *
            / *    y:     * \
         B *- - - - + - - - -* C
            *       :   x   *
              *     :     *
                  * * *

    The center of the sphere is O.
    \Delta ABC is the side view of the cone.
    Note that: OA = OB = OC = 3.

    The volume of a cone is: . V \;= \;\frac{1}{3}\pi r^2h [1]

    In the diagram, we see that: x  =\text{ radius, and }y + 3  =\text{  height}

    So we have: . V \;= \;\frac{1}{3}\pi x^2(y + 3) [2]


    The diagram also tells us that: . x^2 + y^2\:=\:3^2\quad\Rightarrow\quad x^2 \,= \,9 - y^2

    Substitute into [2]: . V \;= \;\frac{1}{3}\pi(9 - y^2)(y + 3)

    Can you finish up?

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  3. #3
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    Thanks

    Thank you so much! This really helped a lot.

    It would be sooooo great if you could finish it up for me, I'm having so much trouble. Please and thank you.
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  4. #4
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    745
    Hello again, philipsach1!

    The rest is "simple" Calculus . . . but okay.


    We have: . V \;= \;\frac{1}{3}\pi(9 - y^2)(y + 3)

    How do we maximize the volume?
    We equate its derivative to zero and solve.

    The function is: . V \;= \;\frac{\pi}{3}\left(-y^3 - 3y^2 + 9y + 27\right)

    So: . V' \;= \;\frac{\pi}{3}\left(-3y^2 - 6y + 9\right) \;=\;-\pi(y^2 + 2y - 3) \;= \;0

    And we have the quadratic: . y^2 + 2y - 3 \;= \;0

    . . which factors: . (y - 1)(y + 3) \;= \;0

    . . and has roots: . y \:= \:1,\;-3


    Since we want positive dimensions: . y \:=\:1\quad\Rightarrow\quad x = 2\sqrt{2}

    . . \begin{array}{ccc}\text{The radius is: } r \:= \:x \:= \:2\sqrt{2} \\ \\ \text{The height is: }h \:= \:y + 3 \:= \:4\end{array}<br /> <br />

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  5. #5
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    Thank you

    Thanks so much! I really appreciate this.
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