A storage bin is to be constructed by removing a sector with central angle $\displaystyle \theta$ from a cicular piece of tin of radius 10 ft and folding the remainder of the tin to form a cone, as shown. What is the maximum volume of a storage bin formed in this fashion.

http://i125.photobucket.com/albums/p...thdraw4624.jpg

I have two questions: is my attempt at modeling the situation correct, and if so, where did I go wrong in solving the model -- I got a negative answer for the angle that produces the mazimum volume.

Let circumference of the material be c1, the circumference of the cone base be c2, and the arclength of the material removed be s.

$\displaystyle c_{1} = 2\pi(10) = 20\pi $

$\displaystyle s = 10\theta$

$\displaystyle c_{2} = 20\pi - 10\theta$

$\displaystyle 2\pi R = 20\pi - 10\theta$

$\displaystyle R = 10 - \frac{5\theta}{\pi}$

$\displaystyle H = \sqrt{10^{2} - \left(\frac{5\theta}{\pi}\right)^{2}} = \sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}}$

$\displaystyle V = \frac{1}{3}\pi R^{2} H$

$\displaystyle V = \frac{\pi}{3}\left(10 - \frac{5\theta}{\pi}\right)^{2}\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}}$

$\displaystyle V = \frac{\pi}{3}\left(100 - \frac{100\theta}{\pi} + \frac{25\theta^{2}}{\pi^{2}}\right)\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}}\longleftarrow\mbox{H ere is the model -- right so far?}$

$\displaystyle \frac{dV}{d\theta} = \frac{\pi}{3}\left[\left(-\frac{100}{\pi} + \frac{50\theta}{\pi^{2}}\right)\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}} + \left(100 - \frac{100\theta}{\pi} + \frac{25\theta^{2}}{\pi^{2}}\right)\frac{-25\theta}{\pi^{2}\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}}}\right]$

$\displaystyle 0 = \left(-\frac{100}{\pi} + \frac{50\theta}{\pi^{2}}\right)\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}} + \left(100 - \frac{100\theta}{\pi} + \frac{25\theta^{2}}{\pi^{2}}\right)\frac{-25\theta}{\pi^{2}\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}}}$

$\displaystyle \left(\frac{100}{\pi} - \frac{50\theta}{\pi^{2}}\right)\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}} = \left(100 - \frac{100\theta}{\pi} + \frac{25\theta^{2}}{\pi^{2}}\right)\frac{-25\theta}{\pi^{2}\sqrt{100 - \frac{25\theta^{2}}{\pi^{2}}}}$

$\displaystyle \pi^{2}\left(\frac{100}{\pi} - \frac{50\theta}{\pi^{2}}\right)\left(100 - \frac{25\theta^{2}}{\pi^{2}}\right) = (-25\theta)\left(100 - \frac{100\theta}{\pi} + 25\theta^{2}\right)$

$\displaystyle 10000\pi - \theta\frac{5000}{\pi} - \theta^{2}\frac{2500}{\pi^{3}}+ \theta^{3}\frac{1250}{\pi^{2}} = -2500\theta + \theta^{2}\frac{2500}{\pi} - 6250\theta^{3}$

$\displaystyle 8\pi^{4} - 4\pi^{2}\theta - 2\theta^{2}+ \pi\theta^{3} = -2\pi^{3}\theta + 2\pi^{2}\theta^{2} - 5\pi^{3}\theta^{3}$

$\displaystyle 0 = \theta^{3}(\pi + 5\pi^{3}) - \theta^{2}(2 + 2\pi^{2}) + \theta(-4\pi^{2} + 2\pi^{3}) + 8\pi^{4}$

The problem is that there is no solution to this in on $\displaystyle 0 \leq \theta \leq 2\pi$ (at least according to my calculator) -- a long way to go for a very disapointing answer. Where did I go wrong?