The model for a profitable account for horseshoe crab farm with harvest rate 250 crabs per month is given by $y' = -\frac{1}{4} + y - \frac{y^2}{a}$, where y has units of kilocrabs. What seems to be the minimum value of $a$ resulting in a stable positive population of crabs? (Find the $a_{min}$ and $y_{min}$.)
I'm confused because trying multiple values of $a$, I can see that the lines of unstable and stable equilibria, at $a = 0.25$, collapse together, and I thought that would be the answer. But apparently, $a=0.25$ is not the answer. I don't have any other ideas though; how might I approach/solve this problem?
Completing the square gives $$ y'=-\frac{1}{4a}(a-4ay+4y^2)=-\frac{1}{4a}(a-a^2+(2y-a)^2) $$ Clearly to get real roots of the right side and thus equilibrium positions one needs $a-a^2<0$ which is the case for $a<0$ and $a>1$. At $a=1$ the double root $y=\frac12$ is not stable, for $a>1$ the root $\frac12(a+\sqrt{a^2-a})$ is positive and a stable equilibrium point.