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Math Help - Related Rates: inverted cone

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    Related Rates: inverted cone

    A cone (pointing downward) has a depth of 8m and a radius of 5m. Beginning at t=0, two things start happening: the radius begins expanding at 1 meter/sec, and water is poured in at a rate of 3 cubic meters/sec. The depth of the cone stays constant.

    At what time does the depth of the water reach a maximum?

    I've got an answer that I feel good about, but I'd like a more elegant method. Specifically, it'd be nice to know if this is possible without rewriting each part as a function in terms of time.

    My answer in white text: t = 5.
    Last edited by Henderson; November 11th 2010 at 02:11 PM. Reason: Found an error in my work.
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    Quote Originally Posted by Henderson View Post
    A cone (pointing downward) has a depth of 8m and a radius of 5m. Beginning at t=0, two things start happening: the radius begins expanding at 1 meter/sec, and water is poured in at a rate of 3 cubic meters/sec. The depth of the cone stays constant.

    At what time does the depth of the water reach a maximum?

    I've got an answer that I feel good about, but I'd like a more elegant method. Specifically, it'd be nice to know if this is possible without rewriting each part as a function in terms of time.

    My answer in white text: t = 5.
    I came up with depth as a function of time also ...

    \displaystyle h = \frac{9t}{\pi(5+t)^2}

    ... and arrived at the same solution. Seems to me that getting depth as a function of time is about as "elegant" as any other method. I'd have to see an alternate approach to judge it "elegance" quotient.
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