Question

prove that the volume of the largest cone that is inscribed in a sphere of radius R is 8/27 of the volume of sphere

Answer 1

the diagram shows a relationship of:
\[r^2 + (h-R)^2 = R^2\]
solve for "h" to plug into Volume equation...
\[h = \sqrt{R^2 - r^2} + R\]
\[V = \frac{\pi}{3} r^2 h\]
\[V = \frac{\pi}{3} r^2 (\sqrt{R^2 - r^2}+R)\]
Next you want to maximize volume of cone, set derivative equal to 0
keeping "R" as a constant
--> derivative using product rule:
\[\frac{dV}{dr} = \frac{2\pi}{3} r (\sqrt{R^2 - r^2} +R) + \frac{\pi}{3} r^2(\frac{-r}{\sqrt{R^2 -r^2}}) = 0\]
solve for "r^2" in terms of R ...
\[r^2 = \frac{8R^2}{9}\]
plug into Volume equation
\[V = \frac{32}{81} \pi R^3\]
Finally divide by volume of sphere to get ratio...
\[\frac{(32/81) \pi R^3}{(4/3) \pi R^3} = \frac{8}{27}\]