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Hello I am trying to solve for the smallest possible radius $r$ for a sphere that can inscribe a circular cylinder of volume 8 cubic units(so the volume is given and is a constant 8 cubic units). I have already solved for a formula which gives the maximum possible volume of a circular cylinder dependent of the radius $r$ of the sphere with

$V_{max of cylinder} = \frac{4\pi r^3}{3\sqrt{3}}$

Returning to the new problem of finding the smallest radius $r$ that can inscribe a circular cylinder of 8 cubic units(but not necessarily a circular cylinder that has an optimal volume which follows the above formula), how can one apply the Arithmetic Geometric Mean Inequality to minimize the radius $r$ of the sphere which inscribes any circular cylinder of volume 8 cubic units? I apologize if the way I stated this problem causes ambiguity, so I will just assume the problem I am really trying to solve is "What is the smallest possible radius $r$ for a sphere that can inscribe a circular cylinder of 8 cubic units?" and that really under these conditions, one may assume that the cylinder itself has an optimal maximum volume following the formula I stated above. For educational purposes, I want to know how to exactly set up problem of applying the Arithmetic Geometric Mean Inequality for minimizing the radius of the sphere which can enclose the circular cylinder of 8 cubic units.

In addition, how would a single-variable calculus approach to finding the minimum radius follow? The formula above I obtained for the maximum possible volume of an inscribed circular cylinder in a sphere for a given sphere radius $r$ was obtained through single variable calculus(the forumla gives the maximum possible volume of a cylinder for a given sphere radius), I want to know if there is a way to similarly apply a single variable calculus approach to finding the minimum radius $r$ that can enclose a known volume for an inscribed circular cylinder. Any help on either questions I have stated would be appreciated, thanks.

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    $\begingroup$ You want the sphere to circumscribe the cylinder, not inscibe it. $\endgroup$ – Ross Millikan Feb 22 at 0:17
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Given your result, you can just set the volume of the cylinder to $8$ and solve for $r$ algebraically. $$8=\frac{4\pi r^3}{3\sqrt 3}\\r^3=\frac {24 \sqrt 3}{4\pi}\\r=\frac {2\sqrt 3}{(4\pi)^{1/3}}$$

If you didn't have that result you would follow the same process you did to achieve it. You might as well start with a unit sphere and find the largest cylinder that fits, then note that the volume will scale with the cube of the radius (as your formula shows) and scale up as required. Let $r$ be the radius and $h$ be the height of the cylinder. Use geometry to get an equation linking $r$ and $h$ so that the cylinder just fits inside the unit sphere. Plug this into the cylinder volume equation so it becomes a function of just one variable, take the derivative, set to zero, and solve.

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