= Solution
For $q>0$, divide the two sink terms by $\mathcal P=C_uSq^2$. Put
$$
a=\frac{C_\rho^2L_\rho N^2}{C_uS},\qquad b=\frac1{L_uC_uS}.
$$
The requested ratio is
$$
\boxed{R(q)=\frac{B+\epsilon}{\mathcal P}=\frac{a}{q}+bq.}
$$
Here $b>0$ and $a\geq0$. With stable stratification $a>0$, $R$ diverges at both $q\downarrow0$ and $q\to\infty$, and has a single minimum at $q_m=\sqrt{a/b}$, with $R_{\min}=2\sqrt{ab}$. Increasing the stable density-gradient magnitude increases $a$, lifting the minimum and moving it rightward. When the <mass density> gradient vanishes, $a=0$ and the curve reduces to the straight line $R=bq$, with zero only as a limiting value at $q=0$ where the original ratio is undefined.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-73-equilibrium.png]
{title=Sink-to-production ratio for zero, subcritical, critical and supercritical density gradients}
The plot uses $\widetilde q=bq=q/(L_uC_uS)$ and $\mathcal A=ab=[L_\rho C_\rho^2/(L_uC_u^2)]\mathrm{Ri}$, so $R=\widetilde q+\mathcal A/\widetilde q$. The horizontal line $R=1$ selects equilibria: two positive intersections for $0<\mathcal A<1/4$, one double intersection at $\mathcal A=1/4$, and none above it. The zero-gradient line has one positive equilibrium; its formal zero root before division by $q$ is the non-turbulent state.
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