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Math Help - Optimization

  1. #1
    Member
    Joined
    Apr 2008
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    191

    Optimization

    Hello!

    Here's a question that I couldn't understand;

    A canvas tent is to be constructed in the shape of a right-circular cone with the ground as base;

    Using the volume V and curved surface area S of the cone,
    V = \frac{1}{3}\pi r^2 h and S = \pi rlFind the dimensions of the cone that maximises the volume for the 4 \sqrt{3 \pi} mē of canvas material, and find this maximum volume.
    My attempt:know the two formulas: V = \frac{1}{3}\pi r^2 h and S = \pi rl, l, of course, is the "slant height". Using the Pythagorean theorem, l^2= r^2+ h^2.
    Since we must have: S= \pi rl= \pi r\sqrt{r^2+ h^2}= 4\sqrt{3\pi}, Squaring both sides of that \pi^2r^2(r^2+ h^2)= 48\pi.Now we must solve for h and replace h in V= \pi r^2h/3 by that to get a problem in just the one variable r. (Then after that we find the derivate to find the largest possible value of that function i.e. maximized.)
    My problem is here: How do we solve for h in \pi^2r^2(r^2+ h^2)= 48\pi??

    Is it: h = \sqrt\frac{{48 \pi - r}}{\pi^2 . r^2} ?

    Thank you,

    Regards
    Last edited by Roam; April 30th 2008 at 01:54 PM.
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