1. **Problem statement:** A cone of the highest possible volume is hollowed out from a wooden cylinder. We need to find the ratio of the volume of the remaining wood to the volume of the cone hollowed out. 2. **Understanding the problem:** The cone with the highest volume that can be inscribed in a cylinder has a specific ratio between its height and radius relative to the cylinder. 3. **Formulas:** - Volume of cylinder: $$V_{cyl} = \pi r^2 h$$ - Volume of cone: $$V_{cone} = \frac{1}{3} \pi r^2 h$$ 4. **Maximizing cone volume inside cylinder:** For a cone inscribed in a cylinder of radius $R$ and height $H$, the cone's radius $r$ and height $h$ satisfy the relation $r = R \left(1 - \frac{h}{H}\right)$ if the cone vertex is at the top of the cylinder. However, the problem usually assumes the cone has the same base radius and height as the cylinder for maximum volume, so $r = R$ and $h = H$. 5. **But the cone of highest volume inside a cylinder is when the cone's height equals the cylinder's height and radius equals the cylinder's radius.** 6. **Calculate volumes:** - Volume of cylinder: $$V_{cyl} = \pi R^2 H$$ - Volume of cone: $$V_{cone} = \frac{1}{3} \pi R^2 H$$ 7. **Volume of remaining wood:** $$V_{wood} = V_{cyl} - V_{cone} = \pi R^2 H - \frac{1}{3} \pi R^2 H = \frac{2}{3} \pi R^2 H$$ 8. **Ratio of volume of remaining wood to volume of cone:** $$\text{Ratio} = \frac{V_{wood}}{V_{cone}} = \frac{\frac{2}{3} \pi R^2 H}{\frac{1}{3} \pi R^2 H} = 2$$ **Final answer:** The ratio of the volume of the remaining wood to the volume of the cone hollowed out is $2$.