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The first question that comes into my mind here is whether any cylinder that touches(at 4 pts) the circumference of the sphere and does not go out of it, has equal volume? Second, how do i mathematically limit the volume of the cylinder to be less than that of a sphere? Squeeze theorem? Please help, thanks! |
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Let $R$ be the radius of the sphere and let $h$ be the height of the cylinder centered on the center of the sphere. By the Pythagorean theorem, the radius of the cylinder is given by $$ r^2 = R^2 - h^2. $$ The volume of the cylinder is hence $$ \begin{align} V &= \pi r^2 h\\ &= \pi (h R^2 - h^3). \end{align} $$ Differentiating with respect to $h$ and equating to $0$ to find extrema gives $$ \frac{dV}{dh}=\pi (R^2 - 3h^2) = 0\\ \therefore h_0 = \frac{R}{\sqrt{3}} $$ The second derivative of the volume with respect to $h$ is always negative such that the volume is maximal at $h = h_0$. Substituting gives $$ V_{max}=\pi \frac{R^3}{3} (1 - \frac{1}{\sqrt{3}}). $$ |
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Hint: In the context of a calculus course, I think you are first expected to argue informally that such a maximal cylinder must have axis that goes through the center of the circle, and that without loss of generality that axis is the $z$-axis. So now suppose that the cylinder meets the $x$-$y$ plane in a circle of radius $t$. Find the height of the cylinder in terms of $t$, and hence the volume. Now use the ordinary tools to maximize. |
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