Answer
$$h = \frac{{4\sqrt 3 }}{3}{\text{ and }}r = \frac{{4\sqrt 2 }}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the image we can calculate }}r{\text{ by using the pythagorean}} \cr
& {\text{theorem}} \cr
& r = \sqrt {{4^2} - {h^2}} \cr
& r = \sqrt {16 - {h^2}} {\text{ }}\left( {\bf{1}} \right) \cr
& {\text{The volume of a cone is:}} \cr
& V = \frac{{\pi {r^2}h}}{3},{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Substitute }}\left( {\bf{1}} \right){\text{ into }}\left( {\bf{2}} \right) \cr
& V = \frac{{\pi {{\left( {\sqrt {16 - {h^2}} } \right)}^2}h}}{3} \cr
& V = \frac{{\pi \left( {16 - {h^2}} \right)h}}{3} \cr
& V = \frac{{16\pi h - \pi {h^3}}}{3} \cr
& V = \frac{{16\pi }}{3}h - \frac{\pi }{3}{h^3} \cr
& {\text{Differentiate both sides with respect to }}h \cr
& \frac{{dV}}{{dh}} = \frac{{16\pi }}{3} - \pi {h^2} \cr
& {\text{Let }}\frac{{dV}}{{dh}} = 0 \cr
& \frac{{16\pi }}{3} - \pi {h^2} = 0 \cr
& \pi {h^2} = \frac{{16\pi }}{3} \cr
& {h^2} = \frac{{16}}{3} \cr
& h = \pm \frac{4}{{\sqrt 3 }},{\text{ Considering }}h > 0 \to h = \frac{4}{{\sqrt 3 }} = \frac{4}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr
& h = \frac{{4\sqrt 3 }}{3} \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{dV}}{{dh}} = \frac{{16\pi }}{3} - \pi {h^2} \cr
& \frac{{{d^2}V}}{{d{h^2}}} = - 2\pi h \cr
& {\text{Evaluate at }}h = \frac{{4\sqrt 3 }}{3} \cr
& \frac{{{d^2}V}}{{d{h^2}}} = - 2\pi \left( {\frac{{4\sqrt 3 }}{3}} \right) < 0,{\text{ then there is a maximum at }}h = \frac{{4\sqrt 3 }}{3} \cr
& {\text{Calculate }}r \cr
& r = \sqrt {16 - {h^2}} \cr
& r = \sqrt {16 - {{\left( {\frac{{4\sqrt 3 }}{3}} \right)}^2}} = \sqrt {16 - \frac{{16}}{3}} = \frac{{4\sqrt 2 }}{3} \cr
& {\text{Then,}} \cr
& h = \frac{{4\sqrt 3 }}{3}{\text{ and }}r = \frac{{4\sqrt 2 }}{3} \cr} $$
