 Solution

Treat the liposome using the usual incompressible-tail packing model of a lipid bilayer: neglect the head-group thickness and let $v$ be the volume filling the hydrophobic part of the lipid vesicle. Both membrane leaflets have area $a$ per head group. Put $d=v/a$, and let $R$ be the unknown outer radius. Counting molecules from the inner and outer surface areas givesEquating the total molecular volume to the volume of the spherical shell therefore givesTo prove uniqueness in the physical domain $R>r_i>0$, defineIts derivative isAlso $q(r_i)=0$ and $q(R)\to \infty$ as $R\to \infty$. By continuity and strict monotonicity, $q(R)=3d$ has exactly one solution with $R>r_i$. This proves the required uniqueness without choosing among ambiguous branches of a cubic formula.

The maximum tail length is a physical admissibility bound, not automatically the actual leaflet thickness. If $r_m$ is the radius where the two tail regions meet, separate membrane leaflet volume balances giveThe unique candidate must satisfyIn particular $R-r_i\le 2l$ is necessary. The premise that a lipid vesicle has assembled supplies this compatibility: arbitrary positive choices of $r_i,v,a,l$ need not allow any such lipid bilayer. For a very large vesicle the two thicknesses tend to $d$, so $R-r_i\simeq 2v/a$. Setting the thickness equal to $2l$ for every molecule would confuse maximum extension with actual packing. The volume-balance argument and its domain are an example of uniqueness of the packed spherical bilayer radius.