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Math Help - Please Help!

  1. #1
    Newbie Mr. Engineer's Avatar
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    Please Help!

    A sector with central angle \theta is cut from a circle of radius 12 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of \theta so that the volume of the cone is a maximum.
    See attached figure
    thanks
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  2. #2
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    Quote Originally Posted by Mr. Engineer View Post
    A sector with central angle \theta is cut from a circle of radius 12 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of \theta so that the volume of the cone is a maximum.
    See attached figure
    thanks
    There's stuff in this link that should solve-enable you: Math Forum - Ask Dr. Math
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  3. #3
    Newbie Mr. Engineer's Avatar
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    Tell if im in the right direction

    Vcone=\frac{1}{3}\pi r^2 h
    \frac{length of sector}{2\pi}=\frac{\theta}{360}

    length of sector=\frac{2\pi\theta}{360}this would be the circumference of the cone

    then from the example in the previous post, the height would be
    cos\frac{\theta}{2}=\frac{h}{12}
    h=12cos\frac{\theta}{2}

    am i right so far?
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  4. #4
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    Quote Originally Posted by Mr. Engineer View Post
    Vcone=\frac{1}{3}\pi r^2 h
    \frac{length of sector}{2\pi {\color{red}\cdot r}}=\frac{\theta}{360}

    length of sector=\frac{2\pi\theta}{360} {\color{red}\cdot 12} this would be the circumference of the cone

    then from the example in the previous post, the height would be
    cos\frac{\theta}{2}=\frac{h}{12}
    h=12cos\frac{\theta}{2}

    am i right so far?
    You are dealing with two different radi:
    - The radius of the sector which is the slanted line from the tip of the cone to it's edge
    - the radius R of the base circle of the cone.

    You know the length l of the arc of the sector: l = \frac{2\pi\theta}{360} \cdot 12= \frac{\pi \cdot \theta}{15}

    which has the same length as the circumference of the base circle:

     \frac{\pi \cdot \theta}{15} = 2\pi \cdot R~\implies~R=\frac{\theta}{30}

    The slanted line s, the radius of the base cicle R and the height of the cone form a right triangle. Use Pythagorean theorem:

    h^2+R^2=s^2~\implies~h=\sqrt{144-\left(\frac{\theta}{30} \right)^2}

    Plug in the terms of R and h into the equation of the volume. You'll get a function with respect of \theta :

    V(\theta)=\frac13 \cdot \pi \cdot \left(\frac{\theta}{30}\right)^2 \cdot \sqrt{144-\left(\frac{\theta}{30} \right)^2}

    To calculate the maximum volume you have to derive this function and solve for \theta the equation:

    V'(\theta) = 0

    EDIT: Removed my typo in the last equation.
    Last edited by earboth; May 11th 2008 at 12:40 PM.
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