Maximum sustainable yield

1. Equilibrium. $r\left(1-\frac NK\right)=E$ gives $N^*=K\left(1-\dfrac Er\right)>0$ and
$$Y(E)=EK\left(1-\frac Er\right)=KE-\frac{K}rE^2.$$
This equilibrium is stable: the equation can be rewritten $\dfrac{dN}{dt}=(r-E)N\left(1-\dfrac N{N^*}\right)$, a logistic equation with growth rate $r-E>0$ and carrying capacity $N^*$, so every positive solution tends to $N^*$. This is why $Y(E)$ is a yield that can be sustained.

2. Maximum. $Y'(E)=K-\dfrac{2K}rE=0\iff E=\dfrac r2$, a maximum of the concave parabola on $[0,r)$.
- $N^*=\dfrac K2$: the population is kept at half its carrying capacity.
- $Y_{\max}=\dfrac{rK}4$.

Remark. $\dfrac K2$ is also the population at which the natural growth rate $rN\left(1-\frac NK\right)$ is largest (it is the top of that parabola), so harvesting exactly this growth is the most one can sustainably take. For an effort $E\ge r$, $\frac{dN}{dt}=N\left[(r-E)-\frac rKN\right]<0$ for every $N>0$, so the population tends to $0$: in this model, a fishing effort of at least $r$ leads to extinction.