1. **Problem statement:** A cylinder is inscribed in a sphere of radius 8 inches. We need to find the maximum volume of this cylinder. 2. **Known values and variables:** - Radius of the sphere: $R = 8$ inches - Let the height of the cylinder be $h$ inches - Let the radius of the cylinder's base be $r$ inches 3. **Relationship between $r$, $h$, and $R$:** Since the cylinder is inscribed in the sphere, the diagonal from the center of the cylinder's base to the top edge lies on the sphere's radius. Using the Pythagorean theorem in the cross-section: $$r^2 + \left(\frac{h}{2}\right)^2 = R^2$$ 4. **Volume formula for the cylinder:** $$V = \pi r^2 h$$ 5. **Express $r^2$ in terms of $h$:** $$r^2 = R^2 - \left(\frac{h}{2}\right)^2 = 64 - \frac{h^2}{4}$$ 6. **Substitute $r^2$ into volume formula:** $$V = \pi \left(64 - \frac{h^2}{4}\right) h = \pi \left(64h - \frac{h^3}{4}\right)$$ 7. **Maximize $V$ by finding critical points:** Take derivative with respect to $h$: $$\frac{dV}{dh} = \pi \left(64 - \frac{3h^2}{4}\right)$$ Set derivative equal to zero: $$64 - \frac{3h^2}{4} = 0$$ 8. **Solve for $h$:** $$64 = \frac{3h^2}{4}$$ $$64 \times 4 = 3h^2$$ $$256 = 3h^2$$ $$h^2 = \frac{256}{3}$$ $$h = \sqrt{\frac{256}{3}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3} \approx 9.2376$$ 9. **Calculate $r$ using $h$:** $$r^2 = 64 - \frac{h^2}{4} = 64 - \frac{256/3}{4} = 64 - \frac{256}{12} = 64 - \frac{64}{3} = \frac{192}{3} - \frac{64}{3} = \frac{128}{3}$$ $$r = \sqrt{\frac{128}{3}} = \frac{8\sqrt{6}}{3} \approx 6.5329$$ 10. **Calculate maximum volume $V_{max}$:** $$V_{max} = \pi r^2 h = \pi \times \frac{128}{3} \times \frac{16\sqrt{3}}{3} = \pi \times \frac{2048\sqrt{3}}{9} \approx 1237.44$$ **Final answer:** The maximum volume of the cylinder is about **1237.44** cubic inches.