Question 1048655
<font face="Times New Roman" size="+2">


Let *[tex \Large x] represent the radius of the smaller container.  Then the radius of the larger container is *[tex \Large 2x].  Let *[tex \Large h] represent the height of the smaller container and then the height of the larger container is *[tex \Large \frac{2}{3}h]


The volume of the larger container is then *[tex \Large \pi(2x)^2\,\cdot\,\frac{2}{3}h\ =\ 4\pi x^2\,\cdot\,\frac{2}{3}h].


The volume of the smaller container is *[tex \Large \pi x^2h], and the volume of 2 of the smaller containers must be *[tex \Large 2\pi x^2h]


Divide the volume of the larger by the volume of the two smaller containers.  If you get a value larger than 1, it is a good deal.  If you get a value less than 1, you made a bad choice.  Equal to 1 is a wash.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4\pi x^2\,\cdot\,\frac{2}{3}h}{2\pi x^2h}]


Notice that all of the variables and *[tex \Large \pi] cancel.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>