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I am trying to find the maximum volume that a cylinder inscribed in an ooctahedron of edge 1 cm can have. Given that the cylinder is on a diagonal of the octahedron. With some calculus and geometry we know that:

The volume of a cylinder is $\pi r^2 h$

In this case $r\leq \sqrt{\frac{3}{2}}$ and $h\leq \sqrt{2}$

From here I think I should come with an optimization problem but I am missing it.

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  • $\begingroup$ I think a diagram is needed here to clarify what is meant by a cylinder (3D shape) on the diagonal of an octahedron (2D shape). $\endgroup$
    – Red Five
    Commented May 15 at 10:29
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    $\begingroup$ @RedFive An octahedron is 3d - Octagon is 2d. $\endgroup$
    – Jaap Scherphuis
    Commented May 15 at 10:31
  • $\begingroup$ Oops. Yes, you are correct. $\endgroup$
    – Red Five
    Commented May 15 at 10:34
  • $\begingroup$ Use the fact that the cylinder touches the octahedron face to find a relationship between the $r$ and $h$ of the cylinder. Then you can get an expression of the cylinder volume that uses just one variable, which you can maximise. $\endgroup$
    – Jaap Scherphuis
    Commented May 15 at 10:37

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Let the diagonal length of octahedron be $d$. Since, edge length is 1, $d$ is $\sqrt{2}$. Consider the cylinder base is distance $x$ away on the diagonal from the closest vertex on octahedron. Then, the height of cylinder is $d - 2x$. The base of cylinder(a circle) is inscribed in a square. Since the diagonal and edge of octahedron are at an angle of 45°, half diagonal of the square(in which cylinder base is inscribed) is equal to $x$. Thus, the radius of inscribed cylinder is $\frac{x}{\sqrt{2}}$. The volume of cylinder $V(x) = \frac{\pi}{2} x^2 (\sqrt{2} - 2x)$. By using AM-GM inequality, $x = \sqrt{2} - 2x$ for maximum volume. This implies cylinder height is given by $x = \frac{\sqrt{2}}{3}$. The radius of cylinder is $\frac{1}{3}$.

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  • $\begingroup$ How do you get the lenght of the square where the base of the cylinder is inscribed? $\endgroup$
    – MiguelCG
    Commented May 15 at 12:23
  • $\begingroup$ It would be nice if you show how you get the square diagonal as $2x$. Although it's just basic trigonometry as one can easily derive the angle from vertices to be $45°$, it would make your answer complete $\endgroup$
    – Gwen
    Commented May 15 at 12:35
  • $\begingroup$ Edited the answer to explain the square half diagonal is equal to the distance from vertex. $\endgroup$
    – gsomani
    Commented May 16 at 5:01

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