Given a sphere of fixed radius a. A right circular cone is is to be
found which will enclose the sphere such that the sphere is tangent
to the cone's lateral surface, the sphere is tangent to the cone's
base at its center, and the ratio Ac/As is minimised where Ac is the
cone's surface area (both lateral and base) and As the sphere's
surface area.

To keep solutions uniform let's denote the cone's altitude by h, base
radius by r and slant height by L.

The surface area of a right circular cone is generated by rotating
a right triangle about one of its legs. The lateral surface area is
generated by the hypotenuse and the base by the other side.

Pappus' rule: If a plane curve is rotated about a line in its plane
such that the curve lies on one side of the line, then the area of
the resulting surface of rotation is equal to the product of the
length of the generating curve and the length of the path of the
center of gravity of the curve.

For our problem the center of gravity of both the hypotenuse L
and the leg r with respect to leg h is r/2. Therefore,

Ac = (L + r)*2π*(r/2) = πr(L+r)
(1)

The leg h is called the axis of the cone. The intersection of the
cone and any plane containing the axis is an isosceles triangle.
Where the length of the two equal sides is L and the base is 2r.
The intersection of the sphere with the plane is the incircle of
of this triangle. The axis of the triangle bisects the apex angle
and the bisector of either of the other two angles intersects
the axis at the center of the incircle. Therefore,

(h-a)/a = L/r
(2)

Solving (2) for L gives

L = r(h-a)/a
(3)

Using this L and the equation L^{2} = h^{2} + r^{2} gives

h = 2ar^{2}/(r^{2}-a^{2})
(4)

Using this h and equation (3) gives

L = r(r^{2}+a^{2})/(r^{2}-a^{2})
(5)

Using this L and equation (1) gives

Ac = 2πr^{4}/(r^{2}-a^{2})
(6)

Since the sphere is of fixed radius a, As is also fixed.
Therefore,