The tricky part is to find the nearest point on a meridian to a given point P. Let's fix the meridian M at φ=0, to keep the algebra simple.
Suppose P has spherical coordinates (θ, φ), with 0 ≤ θ ≤ π and -π < φ ≤ π. Let C be the great circle through P which is perpendicular to M; then we are looking for an intersection of C and M. (There are two such intersections; in the end we will simply choose the one that is on the same side of the equator as P).
The normal of C meets the sphere on the meridian M, at A = (ρ, 0), say. P is on C, so OP is perpendicular to OA, where O is the centre of the sphere. In cartesian coordinates, with r = 1 for simplicity:
A = (sin ρ, 0, cos ρ)
P = (cos φ sin θ, sin φ sin θ, cos θ)
The scalar product is
A.P = sin ρ cos φ sin θ + cos ρ cos θ
This must be zero, giving
tan ρ = -1/(cos φ tan θ)
The required point, Q say, is on M and perpendicular to A, so if Q = (σ, 0) in spherical coordinates, we have
tan σ = -1/tan ρ = cos φ tan θ
So
Q = (arctan (cos φ tan θ), 0)
If this Q is on the wrong side of the equator, take
Q = (π - arctan (cos φ tan θ), π)
You might check out work by George Polya. In one of his "popular" books (I think Induction and analogy in mathematics) he defines the Isoperimetric Quotient (IQ) as volume squared over surface area cubed (perhaps times $36\pi$ to get a sphere to one). Amazingly, not all the platonic solids are optimum for their number of faces. That is at a simple level but a nice starting place. It looks like his book Isoperimetric inequalities in mathematical physics with Gábor Szegő addresses these things at a higher level. This is 60 years old but his ideas and exposition are famous. You could search for isoperimetric quotient (although you'll find stuff for plane curves) and also for references to Polya.
Best Answer
There is considerable literature on this question, and closely related variations. See:
(Image from Paul Sutcliffe.)
According to
the Tammes problem is solved exactly for
Fig.1 from Musin & Tarasov: $n=14$.
Added (8Sep15): The exact radius for $n=10$ was just found:
Fig.1b from Sugimoto & Tanemura.
Added (31Dec2017) in response to a question by @R_Berger: For $n=20$, the best arrangement for the Tammes problem is not the dodecahedron's vertices. The optimal is unknown, but this beats the dodecahedron:
Coordinates from Neil Sloane link, due to R.H. Hardin, N.J.A. Sloane & W.D. Smith (1994).