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August 30, 2014

August 30, 2014

Posted by **Ken** on Monday, November 12, 2012 at 1:11am.

- Calculus I -
**Steve**, Monday, November 12, 2012 at 4:02amthe volume of water when the marble is just covered is

v = pi * 4^2 * 2r - 4/3 pi r^3

= 32pi r - 4/3 pi r^3

dv/dr = 32pi - 4pi r^2

dv/dr = 0 when r = √8

since d2v/dr2 < 0, this is a max.

- Calculus I -
**Ken**, Monday, November 12, 2012 at 8:46amThank you! Question: how did you reach d2v/dr2? (did you take a second derivative?)

- Calculus I -
**Steve**, Monday, November 12, 2012 at 9:50amthat's what it is. It is used to determine whether an extremum is a max or min.

- Calculus I -
**Ian**, Monday, November 12, 2012 at 11:18amHow did you get 2r for the height in the volume of water?

- Calculus I -
**Steve**, Monday, November 12, 2012 at 12:38pmcome on, guy! the diameter of the ball is the depth of the water!

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