Example 4 dealt with the case 4h > kM2 in the equation dx/dt = kx(M − x) = h that descr...

Example 4 dealt with the case 4h > kM2 in the equation dx/dt = kx(M − x) = h that describes constant-rate harvesting of a logistic population. Problems deal with the other cases.

If 4h = kM2, show that typical solution curves look as illustrated in Fig. Thus if x0 M/2, then

then

x(t) = 0 after a finite period of time, so the lake is fished out. The critical point x = M/2 might be called semistable, because it looks stable from one side, unstable from the other.