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 \(36\)?

FeedMePi Jun 14, 2021

#1**+3 **

So for this question, formulas for volume and surface areas of spheres are essential. The volume of P is defined as \(\frac{4}{3}πr^3 = \frac{4}{3}π36^3\), and Surface area of P is \(4πr^2\), which in this case is equivalent to \(4π36^2\). Once we put this into the inequality (volume of P)/(surface area of P) ≤ t, we have, after simplification 12≤t. This, the smallest real number t should be 12.

EnchantedLava68 Jun 14, 2021