Regular pyramid and sphere

Let the center of the sphere be $O$ , the top vertex of the pyramid be $A$ , one corner of the square base be $B$ , and the center of the square be $C$ . Let the height of the pyramid be $AC = h$ , and the distance between the center of the sphere and the center of the square be $OC = p$ , the radius of the sphere be $OA = OB = r$ , and the sides of the square be $s$ .

Then $BC = \frac{\sqrt{2}}{2}s$ , and by the Pythagorean Theorem on $\triangle OCB$ , $p^2 + \frac{1}{2}s^2 = r^2$ .

The volume of the pyramid is $V = \frac{1}{3}s^2h = 8\sqrt{3}$ . By segment addition on $AC$ , $p + r = h$ , so $\frac{1}{3}s^2(p + r) = 8\sqrt{3}$ .

Combining $p^2 + \frac{1}{2}s^2 = r^2$ and $\frac{1}{3}s^2(p + r) = 8\sqrt{3}$ to eliminate $s^2$ and rearranging gives $r^3 + pr^2 - p^2r - p^3 = 12\sqrt{3}$ .

By implicit differentiation on $r^3 + pr^2 - p^2r - p^3 = 12\sqrt{3}$ , $3r^2 \cdot \frac{dr}{dp} + 2pr \cdot \frac{dr}{dp} + r^2 - p^2 \cdot \frac{dr}{dp} - 2pr - 3p^2 = 0$ . Since the minimum surface area of the sphere occurs with the minimum $r$ , which is when $\frac{dr}{dp} = 0$ (since $\frac{dr^2}{d^2p} > 0$ for $p > 0$ ), this can be substituted in to give $r^2 - 2pr - 3p^2 = 0$ or $(r - 3p)(r + p) = 0$ . Since $r > 0$ and $p > 0$ , that means $r + p \neq 0$ , which means $r - 3p = 0$ or $p = \frac{1}{3}r$ . Substituting $p = \frac{1}{3}r$ into $r^3 + pr^2 - p^2r - p^3 = 12\sqrt{3}$ and solving gives $r = \frac{3\sqrt{3}}{2}$ , which is the radius of the sphere with the minimum surface area.

The minimum surface area of the sphere is then $S = 4\pi r^2 = 4 \pi (\frac{3\sqrt{3}}{2})^2 = 27\pi$ , so $\lambda = \boxed{27}$ .