A funnel is made up of a partial cone and a cylinder as shown in the figure. The maximum amount of liquid that can be in the funnel at any given time is 16.59375π cubic centimeters. Given this information, what is the volume of the partial cone that makes up the top part of the funnel? A.15.75π cm3 B. 17.25π cm3 C. 16.33π cm3 D. 12.5π cm3

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A funnel is made up of a partial cone and a cylinder as shown in the figure. The maximum amount of liquid that can be in the funnel at any given time is 16.59375π cubic centimeters. Given this information, what is the volume of the partial cone that makes up the top part of the funnel? A.15.75π cm3 B. 17.25π cm3 C. 16.33π cm3 D. 12.5π cm3

Geometry
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Wait, where is the figure?
|dw:1436875930013:dw|

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I am assuming that the cone's tip ( the cone that the partial cone would be a part of) will coincide with the bottom of the cylinder volume when filled = volume of the cone + 2/3 volume of the cylinder 16.59375 pi = (1/3) pi (3)^2 * (h+ 4) + (2/3) pi (0.,75)^2 h where h is the height of the cylinder
3( h + 4) pi + 0.375 pi h = 16.59375pi 3hpi + 12 pi + 0.375 pi h = 16.59375 pi 3h + 12 + 0.375h = 16.59375 h = (16.59375 - 12) / 3.375 h = 1.3611
the volume of the cylinder = pi * (0.75)^2 * 1.3611 volume of partial cone = 16.59375 - volume of cylinder
this gives 15.83pt which is close but not equal to choice A
Perhaps my initial assumption was not valid another way to do this would be to use calculus but i dont know if you are familiar with this
http://keisan.casio.com/exec/system/1223372110 this gives 15.75 pi

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