= Solution
Differentiating $R=a/q+bq$ gives $R'=-a/q^2+b$, so for positive stratification its minimum has $q_m^2=a/b$. At that point
$$
\frac{B}{\mathcal P}=\frac{a}{q_m}=\sqrt{ab},\qquad\frac{\epsilon}{\mathcal P}=bq_m=\sqrt{ab}.
$$
Thus the two sinks are equal. The <buoyancy fraction of total turbulent sinks> is always $B/(B+\epsilon)=1/2$ there. But the <Flux Richardson number> defined in part (i) is instead
$$
\boxed{\mathrm{Ri}_f(q_m)=\sqrt{ab}=\frac{R_{\min}}2.}
$$
It equals $1/2$ \b[only when the minimum is also an equilibrium], $R_{\min}=1$, namely at the critical gradient. This is the qualification needed for the printed assertion; an arbitrary minimum on a curve plotted away from equilibrium need not have $\mathrm{Ri}_f=1/2$.
For example choose closure parameters $C_u=C_\rho=1/4$, units with $L_u=L_\rho=S=1$, and $N^2=1/16$. Then $a=1/64$, $b=4$, $q_m=1/16$ and $R_{\min}=1/2$. The actual flux Richardson number at this minimum is $1/4$, while the <buoyancy> fraction of the total sinks is $1/2$. This is a concrete counterexample to the unqualified wording. With zero <mass density> gradient there is no positive interior minimum at all, and $\mathrm{Ri}_f=0$.
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