1. **Problem statement:** Find the volume of the greatest cylinder that can be inscribed in a cone with height $h$ and semi-vertical angle $30^\circ$. The goal is to show that this volume is $\frac{4}{81} \pi h^3$. 2. **Understanding the problem:** The cone has height $h$ and semi-vertical angle $30^\circ$. The cylinder is inscribed inside the cone, meaning it fits perfectly inside with its base on the cone's base and its top inside the cone. 3. **Key formulas:** - Volume of a cylinder: $$V = \pi r^2 H$$ where $r$ is the radius and $H$ is the height of the cylinder. - Relationship between radius and height in the cone due to the angle: The radius of the cone at height $x$ from the apex is $$r_{cone}(x) = x \tan 30^\circ = \frac{x}{\sqrt{3}}$$. 4. **Setting variables:** - Let the height of the cylinder be $H$. - The cylinder is placed inside the cone such that its top is at height $h - H$ from the apex. - The radius of the cylinder $r$ equals the radius of the cone at height $h - H$, so $$r = (h - H) \tan 30^\circ = \frac{h - H}{\sqrt{3}}$$. 5. **Volume of the cylinder:** $$V = \pi r^2 H = \pi \left(\frac{h - H}{\sqrt{3}}\right)^2 H = \pi \frac{(h - H)^2}{3} H = \frac{\pi}{3} (h - H)^2 H$$ 6. **Maximizing the volume:** - To find the maximum volume, differentiate $V$ with respect to $H$ and set to zero: $$\frac{dV}{dH} = \frac{\pi}{3} \left[ 2(h - H)(-1) H + (h - H)^2 \right] = 0$$ - Simplify: $$\frac{\pi}{3} \left[ -2H(h - H) + (h - H)^2 \right] = 0$$ - Divide both sides by $\frac{\pi}{3}$: $$-2H(h - H) + (h - H)^2 = 0$$ - Expand: $$-2Hh + 2H^2 + h^2 - 2hH + H^2 = 0$$ - Combine like terms: $$h^2 - 4hH + 3H^2 = 0$$ 7. **Solve quadratic in $H$:** $$3H^2 - 4hH + h^2 = 0$$ - Using quadratic formula: $$H = \frac{4h \pm \sqrt{(4h)^2 - 4 \cdot 3 \cdot h^2}}{2 \cdot 3} = \frac{4h \pm \sqrt{16h^2 - 12h^2}}{6} = \frac{4h \pm 2h}{6}$$ - Possible solutions: $$H = \frac{4h + 2h}{6} = h$$ $$H = \frac{4h - 2h}{6} = \frac{2h}{6} = \frac{h}{3}$$ - $H = h$ corresponds to zero radius, so discard. - Maximum volume at $H = \frac{h}{3}$. 8. **Calculate maximum volume:** $$r = \frac{h - H}{\sqrt{3}} = \frac{h - \frac{h}{3}}{\sqrt{3}} = \frac{\frac{2h}{3}}{\sqrt{3}} = \frac{2h}{3\sqrt{3}}$$ $$V = \pi r^2 H = \pi \left(\frac{2h}{3\sqrt{3}}\right)^2 \cdot \frac{h}{3} = \pi \frac{4h^2}{9 \cdot 3} \cdot \frac{h}{3} = \pi \frac{4h^3}{81} = \frac{4}{81} \pi h^3$$ **Final answer:** The volume of the greatest cylinder inscribed in the cone is $$\boxed{\frac{4}{81} \pi h^3}$$.