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        riddles >> medium >> Cocktail Glass
        
(Message started by: ThudanBlunder on Jan 23^rd, 2009, 12:53am)

Title: Cocktail Glass
Post by ThudanBlunder on Jan 23^rd, 2009, 12:53am

If a full cocktail glass has dimensions as below, what size sphere would we need to place into it to displace the maximum amount of fluid?

Title: Re: Cocktail Glass
Post by balakrishnan on Jan 23^rd, 2009, 9:06am

The radius for which the sphere grazes the cocktail container at its edge would be h cot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) and the radius at which the sphere is just completely immersed into the container is h/(1+sec(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)), implying the radius of the sphere must be
h/(1+sec(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)) <=R <= h cot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) . If R is the radius of the sphere, we can see that (using simple calculus) the volume displaced is given by
V_d=2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifR³/3 -http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifR² (R sec(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)-h)+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif(R sec(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)-h)³/3
Differentiating w.r.t. R and equating to zero , we get
R=hcos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/[(1-cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif))(1+2cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)] or R=h/(sec(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)-1)
However R=h/(sec(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)-1) cannot be true since R> h cot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)
Also R=hcos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/[(1-cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif))(1+2cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)] is true only when http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif>=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4
Otherwise
R=h cot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)

So answer is
R=hcos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/[(1-cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif))(1+2cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)] when http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif>=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4
R=h cot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif) when http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif<=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4

Title: Re: Cocktail Glass
Post by Immanuel_Bonfils on Mar 4^th, 2009, 12:15pm

The upper limit of R for maximum volume is h.cothttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif.cschttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif, lower than
h/(sechttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif-1) but higher than h.coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif/[(1-coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)(1+2coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)] for any 0<http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif<http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2 .
So let's drink it before it'll be all spilled out.