An open, right-circular cone of radius $r$ and height $h$ has a volume of $36\pi {\text{m}}^3.$

1. Show that the surface area, $A,$ satisfies

   $$A^2=k{\pi }^2(k{h}^{-2}+h)$$

   where $k\in \mathbb{N}$ is a number to be determined.

2. Hence, find the smallest possible value for the surface area of the exterior of the cone, giving your answer in the form

   $$k\pi \sqrt{3},k\in \mathbb{N}$$

The volume of a cone is

The surface area of the curved part of the cone is

where $s$ is the slant height:

Since the cone is open, there is no circular base.

The slant height is the hypotenuse of this right-angled triangle:

As $A$ increases, $A^2$ increases, so to find the $h$ which minimises $A$ you can find the $h$ which minimises $A^2.$ In other words, solve

Give the value of $k$.