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Math Help - Maximum volume of cone

  1. #1
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    Maximum volume of cone

    A right-angled triangle ABC with hypotenuse AB of length 4metres is revolved about BC to generate a right circular cone. Show that the maximum volume of the cone is obtained when the ratio BC:CA is 1:\sqrt{2}
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Let h = BC and r = AC

    You get the equation: h^2 + r^2 = 4^2

    The volume of a cone is then:

    V = \frac13\pi r^2 h

    Max volume is given by the derivative of the volume and set to 0. You have to use a substitution so that you have the volume in terms of r only, or in terms of h only.

    You will have a value of r or h, depending on what you did and you can substitute that in the first equation to get the value of the other unknown, after which, you should get the ratio asked for.

    EDIT: Really silly mistakes
    Last edited by Unknown008; June 13th 2011 at 01:46 AM.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    Let r = BC and h = AC

    You get the equation: h^2 + r^2 = 4

    The volume of a cone is then:

    V = \frac23\pi r^2 h

    Max volume is given by the derivative of the volume and set to 0. You have to use a substitution so that you have the volume in terms of r only, or in terms of h only.

    You will have a value of r or h, depending on what you did and you can substitute that in the first equation to get the value of the other unknown, after which, you should get the ratio asked for.
    Let BC be r and AC be h
    V=\frac{1}{3}{\pi}r^2h

    r^2+h^2=4^2
    r^2=16-h^2

    V=\frac{1}{3}{\pi}(16-h^2)h
    \frac{dV}{dh}=\frac{h}{3}{\pi}(-2h)+(16-h^2)(\frac{1}{3}{\pi})

    \frac{dV}{dh}=0,

    \frac{h}{3}{\pi}(-2h)+(16-h^2)(\frac{1}{3}{\pi})=0
    -\frac{2}{3}{\pi}h^2+\frac{16}{3}{\pi}-\frac{h^2}{3}{\pi}=0
    (\frac{2}{3}{\pi}+\frac{{\pi}}{3})h^2=\frac{16}{3}  {\pi}
    h=2.309

    sub h=2.309, r=3.266

    \frac{r}{h}=BC:CA=1:1.414=1:\sqrt{2} (shown)
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oops, I just realised I gave you the wrong formula for the volume and the pythagoras' theorem. And... BC was supposed to be h and AC be r. My bad, I'm losing my touch

    You could have kept it in exact form here:

    \left(\frac23 \pi + \frac{\pi}{3}\right)h^2 = \frac{16}{3}\pi

    \pi h^2 = \frac{16}{3}\pi

    h^2 = \frac{16}{3}

    h = \frac{4}{\sqrt{3}}

    Then r = \sqrt{\frac{48-16}{3}} = \frac{4\sqrt2}{\sqrt3}

    Then the ratio = AC:BC = \frac{4\sqrt2}{\sqrt3} : \frac{4}{\sqrt{3}} = \sqrt2 : 1
    Last edited by Unknown008; June 13th 2011 at 02:05 AM.
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