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 of diagonal length $2x$. 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 $\frac{1}{3}$.

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 $\frac{1}{3}$.

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 of diagonal length $2x$. 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 $x = \frac{\sqrt{2}}{3}$

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 of diagonal length $2x$. 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 $\frac{1}{3}$.