The volume of the largest cone that can be inscribed in a sphere of radius $9$ cm is
- $\displaystyle \frac{32}{3}\pi \ \mathrm{cc}$
- $72\pi \ \mathrm{cc}$
- $288\pi \ \mathrm{cc}$
- $\displaystyle \frac{288}{3}\pi \ \mathrm{cc}$
For a cone of height h and base radius r inscribed in a sphere of radius R=9, the relationship is r^2 = R^2 - (h-R)^2. Substituting R=9, r^2 = 81 - (h-9)^2 = 18h - h^2. The volume V = (1/3)pi(r^2)h = (1/3)pi(18h^2 - h^3). Setting the derivative dV/dh = 0 gives 36h - 3h^2 = 0, so h=12. Then r^2 = 18(12) - 144 = 72. Volume = (1/3)pi(72)(12) = 288pi.
The largest cone has its vertex and base on the surface of the sphere, so using the volume formula, one third times pi times base radius squared times height, we maximize it with the base radius squared plus the height squared relation. Substituting the sphere radius of 9, the maximum height is 12 cm and the base radius is 6 times the square root of 2 cm. The maximum volume is one third times pi times 72 times 12, which is 288 pi cc.