Polyhedron $P$ is inscribed in a sphere of radius $36$ (meaning that all vertices of $P$ lie on the sphere surface). What is the least upper bound on the ratio $$\frac{\text{volume of }P}{\text{surface area of }P}~?$$In other words, what is the smallest real number $t$ such that $$\frac{\text{volume of }P}{\text{surface area of }P} \le t$$must be true for all polyhedra $P$ that can be inscribed in a sphere of radius $36$?
Now I want to maximize this ratio, which means I want to maximize the inradius of $P$. The inradius of $P$ is at most the radius of the sphere that circumscribes $P$, which is given as $36$. So the inradius of $P$ can be anything from $0$ to $36$, and the ratio can be anything from $0$ to $12$. The least upper bound of this ratio is then $12$, since any larger value would not be an upper bound for all possible $P$. Therefore, the smallest $t$ that satisfies the inequality is $t = 12$. 