A marble in a pot

Question

I have been sitting here staring at this problem for hours and I have no idea where to start..... the problem reads:

Imagine a flat bottomed cylindrical pot with a radius of $4$. A marble with a radius $0 < r < 4$ is placed in the bottom of the pot. What is the radius of the marble that requires the most water to cover it completely?

I really hope somebody can help me, please!

Answer 1

The volume of water needed to cover a marble of radius $r$ is te volume of the cylinder of water minus de volume of the marble, this is $V(r) = \pi(4)^2(2r)-\frac{4}{3}\pi r^3 = 32\pi r - \frac{4}{3}\pi r^3$.

We want the maximum of $V$, since $\displaystyle\frac{dV}{dr} = 32\pi - 4\pi r^2$ we have $\displaystyle\frac{dV}{dr} = 0$ if $32\pi - 4\pi r^2=0$, then $r=\sqrt{8}=2\sqrt{2}$. You should verify that $2\sqrt{2}$ is a maximum, this is, show that $V''(2\sqrt{2})<0$.

Answer 2

I presume you know a little calculus? If so then find the maximum of Vol(water) = Vol(cylinder) - Vol(sphere), where the depth of the cylinder is the same as the diameter of the sphere.