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Topic: Cocktail Glass (Read 718 times) |
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ThudnBlunder
wu::riddles Moderator
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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?
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balakrishnan
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Re: Cocktail Glass
« Reply #1 on: Jan 23rd, 2009, 9:06am » |
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The radius for which the sphere grazes the cocktail container at its edge would be h cot( ) and the radius at which the sphere is just completely immersed into the container is h/(1+sec()), implying the radius of the sphere must be
h/(1+sec()) <=R <= h cot() . If R is the radius of the sphere, we can see that (using simple calculus) the volume displaced is given by
Vd=2 R3/3 -R2 (R sec()-h)+(R sec()-h)3/3
Differentiating w.r.t. R and equating to zero , we get
R=hcos()/[(1-cos())(1+2cos()] or R=h/(sec()-1)
However R=h/(sec()-1) cannot be true since R> h cot()
Also R=hcos()/[(1-cos())(1+2cos()] is true only when >=/4
Otherwise
R=h cot()
So answer is
R=hcos()/[(1-cos())(1+2cos()] when >=/4
R=h cot() when <=/4
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| « Last Edit: Jan 23rd, 2009, 1:08pm by balakrishnan » |
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Immanuel_Bonfils
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Re: Cocktail Glass
« Reply #2 on: Mar 4th, 2009, 12:15pm » |
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The upper limit of R for maximum volume is h.cot.csc, lower than
h/(sec-1) but higher than h.cos/[(1-cos)(1+2cos)] for any 0<</2 .
So let's drink it before it'll be all spilled out.
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