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Math Help - Inscribed Figues

  1. #1
    Junior Member Freaky-Person's Avatar
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    Inscribed Figues

    We had this question on a test today, and I spent an actual hour after school sitting and staring it.

    I tried both Pythagorean theorem and similar triangles, and all I've achieved is creating more roots than a forest, and adding a few new variables.

    Find the greatest curved surface area of a right circular cone that can be inscribed in a sphere of a fixed radius R.

    Please help... T.T
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  2. #2
    Eater of Worlds
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    These mostly ask for the greatest volume. This is a change.

    Let r=radius of cone, R=radius of sphere, h=height of cone.

    \text{surface area of cone}={\pi}r\sqrt{r^{2}+h^{2}}

    By using triangles, we can see that h=y+R

    Where, by triangle OAB, y=\sqrt{R^{2}-r^{2}}

    So, h=\sqrt{R^{2}-r^{2}}+R

    Now, R is a constant, so we have it down to one variable, r, the radius of the cone.

    S={\pi}r\sqrt{r^{2}+(\sqrt{R^{2}-r^{2}}+R)^{2}}

    This produces a booger to differentiate, though.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by galactus View Post
    These mostly ask for the greatest volume. This is a change.

    Let r=radius of cone, R=radius of sphere, h=height of cone.

    \text{surface area of cone}={\pi}r\sqrt{r^{2}+h^{2}}

    By using triangles, we can see that h=y+R

    Where, by triangle OAB, y=\sqrt{R^{2}-r^{2}}

    So, h=\sqrt{R^{2}-r^{2}}+R

    Now, R is a constant, so we have it down to one variable, r, the radius of the cone.

    S={\pi}r\sqrt{r^{2}+(\sqrt{R^{2}-r^{2}}+R)^{2}}

    This produces a booger to differentiate, though.

    Doing this, I get r\approx{0.85R}, \;\ or \;\ .525R

    Have to check to see which gives the max.
    I did that, but there is no way a grade 12 university calculus class would be forced to deal with that many radicals in 1 hour and 20 minutes.
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  4. #4
    Eater of Worlds
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    Then, we'll try another way. Write it in terms of theta using trig expressions. That'll probably be just as messy. The formula for the surface area of a cone has that darn radical in it.
    Last edited by galactus; May 8th 2008 at 03:56 PM.
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  5. #5
    Eater of Worlds
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    One thing you could try is to maximize the volume. Is it safe to say the surface area will be maximized also?.

    Why can't you use technology to solve it. That's what I did. I made a mistake on those solutions I posted. That is why I deleted them.
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  6. #6
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by galactus View Post
    One thing you could try is to maximize the volume. Is it safe to say the surface area will be maximized also?.

    Why can't you use technology to solve it. That's what I did. I made a mistake on those solutions I posted. That is why I deleted them.
    It would be safe, except it's not. I have no proof of it being so because volume also depends on height, and though height and 's' correlate, there's probably still a problem with it.

    Sides we didn't prove that.

    And if you know such technology, feel free to share.
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  7. #7
    Eater of Worlds
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    I just use my TI-92 to differentiate and solve tough problems. That's all I meant.

    I got it now I beleive. I made an error before. I get a radius of r=\frac{2\sqrt{2}R}{3}
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  8. #8
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by galactus View Post
    I just use my TI-92 to differentiate and solve tough problems. That's all I meant.
    oh, well that still doesn't explain how we were supposed to do that on paper... T.T
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  9. #9
    Eater of Worlds
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    You're probably curious. Here's the derivative I got with the calculator.

    \frac{{\pi}((3r+2R)\sqrt{(R-r)R}R-(3r-2R)\sqrt{(r+R)R}R}{2\sqrt{(r+R)R}\sqrt{(R-r)R}}

    I set this equal to 0 and solved for r. That gave the solution of

    r=\frac{2\sqrt{2}R}{3}, \;\ h=\frac{R}{3}+R

    I am sorry about the by hand differentiation. I certainly am not going to do it. That is what calculators are for. At least, when they are this onerous.

    I am confident with the solution I just provided. Gives you something to shoot for if you must do it by hand. Maybe someone else with a clever approach will post.

    Good luck
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