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Thread: Calculus-Optimization

  1. #1
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    Calculus-Optimization

    Ok, this problem is killing me, can I get some help?

    Find the volume of the largest(maximize) right circular cone that can be inscribed in a sphere of radius 4.



    So, so far ive gotten $\displaystyle x^2 + y^2 = 4^2$ so $\displaystyle y^2 = 4^2 + x^2$ the $\displaystyle y= sqrroot 16-x^2$ and therfore $\displaystyle h=4+y$ so $\displaystyle h=4+ sqrroot 16-x^2$

    Am I going in the right direction?
    Thanks!!
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  2. #2
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    Well, then I plugged h into teh volume formula, so I had:

    $\displaystyle V=(1/3)pi r^2 (4+ \sqrt{16-x^2})$

    Then I have to find the derivative of it, but that is what is driving me nuts.

    this was the deriv I got, but how in the world do I simplify this thing? (My algebra isnt that great im finding out.. )
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  3. #3
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    $\displaystyle V = \frac{\pi}{3} x^2(y+4)$

    $\displaystyle V = \frac{\pi}{3}(x^2y+4x^2)$

    $\displaystyle \frac{dV}{dx} = \frac{\pi}{3}(x^2 \frac{dy}{dx} + 2xy + 8x)$

    setting $\displaystyle \frac{dV}{dx} = 0$ ...

    $\displaystyle x \frac{dy}{dx} + 2y + 8 = 0$

    since $\displaystyle \frac{dy}{dx} = -\frac{x}{y}$ ...

    $\displaystyle -\frac{x^2}{y} + 2y + 8 = 0$

    $\displaystyle -x^2 + 2y^2 + 8y = 0$

    $\displaystyle (y^2 - 16) + 2y^2 + 8y = 0$

    $\displaystyle 3y^2 + 8y - 16 = 0$

    $\displaystyle (y+4)(3y-4) = 0$

    $\displaystyle y = \frac{4}{3}$

    $\displaystyle x = \sqrt{16 - \frac{16}{9}} = \frac{8\sqrt{2}}{3}$

    $\displaystyle V_{max} = \frac{2048\pi}{81}$
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  4. #4
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    now that I think about it, this may be the easier way to go ...

    $\displaystyle V = \frac{\pi}{3}x^2(y+4)$

    $\displaystyle V = \frac{\pi}{3}(16 - y^2)(y+4)$

    $\displaystyle V = \frac{\pi}{3}(64 + 16y - 4y^2 - y^3)$

    $\displaystyle \frac{dV}{dy} = \frac{\pi}{3}(16 - 8y - 3y^2)$

    $\displaystyle \frac{\pi}{3}(16 - 8y - 3y^2) = 0$

    $\displaystyle \frac{\pi}{3}(4+y)(4 - 3y) = 0$

    $\displaystyle y = \frac{4}{3}$ ... and so on.
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  5. #5
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    hmmm...ok, so when I get $\displaystyle y= 4/3$ (since we cant use the -4, cause that just wont work) I plug y back into the $\displaystyle V=(pi/3) x^2 (4+y) $ so I get $\displaystyle (pi/3)x^2(16/3)$ ??

    This problem...sheew, im not liking it ..grrrr
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