= 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> gives
$$
M_i=\frac{4\pi r_i^2}{a},\qquad M_o=\frac{4\pi R^2}{a}.
$$
Equating the total molecular volume to the volume of the <spherical shell> therefore gives
$$
\frac{4\pi}{3}(R^3-r_i^3)=v(M_i+M_o),\qquad
\boxed{R^3-r_i^3=3d(R^2+r_i^2).}
$$
To prove uniqueness in the physical domain $R>r_i>0$, define
$$
q(R)=\frac{R^3-r_i^3}{R^2+r_i^2}.
$$
Its <derivative> is
$$
q'(R)=\frac{R^4+3R^2r_i^2+2Rr_i^3}{(R^2+r_i^2)^2}>0.
$$
Also $q(r_i)=0$ and $q(R)\to\infty$ as $R\to\infty$. By <continuity> and strict <monotonicity>, $q(R)=3d$ has \b[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 give
$$
r_m^3=r_i^3+3d r_i^2,\qquad
R^3-r_m^3=3d R^2.
$$
The unique candidate must satisfy
$$
\boxed{r_m-r_i\leq l,\qquad R-r_m\leq l.}
$$
In particular $R-r_i\leq2l$ 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 $R-r_i\simeq2v/a$. 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>.
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