[SOLVED] Modeling problem -- optimization

• Jun 1st 2009, 12:34 PM
sinewave85
[SOLVED] Modeling problem -- optimization
A storage bin is to be constructed by removing a sector with central angle $\theta$ 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.

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

$s = 10\theta$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The problem is that there is no solution to this in on $0 \leq \theta \leq 2\pi$ (at least according to my calculator) -- a long way to go for a very disapointing answer. Where did I go wrong?
• Jun 1st 2009, 04:18 PM
Media_Man
Simple Substitution
The first error I see is on line 6. By your formulation, $R = 10 - \frac{5\theta}{\pi}$ . By Pythagoras, $H^2=10^2-R^2=100-(10 - \frac{5\theta}{\pi})^2$ . Looks like a case of simple oversight. Try correcting this and looking at the model again.
• Jun 1st 2009, 06:27 PM
sinewave85
Quote:

Originally Posted by Media_Man
The first error I see is on line 6. By your formulation, $R = 10 - \frac{5\theta}{\pi}$ . By Pythagoras, $H^2=10^2-R^2=100-(10 - \frac{5\theta}{\pi})^2$ . Looks like a case of simple oversight. Try correcting this and looking at the model again.

Oh, bleeping bleep!!!! I can't belive... oh for the love of Mike. I had notes on this problem on three separate sheets of paper, and somehow I copied it like that onto my final draft, so to speak, before I solved it. I looked over and over the tricky parts of this -- but not the simple parts at the top. It just goes to show the value of a fresh set of eyes when you are so immersed in something. Thanks so much, Media Man. I hope that was the only problem. Not looking forward to redoing all the rest of it, but you gotta do what you gotta do.

Edit: Got it! The maximum volume is about $403 ft^{3}$. Much easier to solve without the error. Thanks again.