Question 999834
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A sphere radius that intersects the edge of the cylinder base is the hypotenuse of a right triangle with legs that are cylinder radius and one-half of the cylinder height.  Given a sphere with radius 6, the radius and height of the cylinder are related thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ =\ 36\ -\ \left(\frac{h}{2}\right)^2]


The volume of the cylinder is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V\ =\ \pi r^2h]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(h)\ =\ \pi\left(36\ -\ \left(\frac{h}{2}\right)^2\right)h]


Simplifying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(h)\ =\ \pi\left(36h\ -\ \frac{h^3}{4}\right)]


Take the first derivative


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V'(h)\ =\ \pi\left(36\ -\ \frac{3h^2}{4}\right)]


Set the first derivative equal to zero and solve for the value of the height that gives the maximum volume.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \