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

Example 4 dealt with the case Ah > 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 x(r) = 0 after a finite period of time, so the lake is fished out (whatever the initial population). [Suggestion: Complete the square to rewrite the differential equation in the form dx/dt = −k\(x − a)2 + b2]. Then solve explicitly by separation of variables.] The results of this and the previous problem (together with Example 4) show that is a critical harvesting rate for a logistic population. At any lesser harvesting rate the population approaches a limiting population N that is less than M(why?), whereas at any greater harvesting rale the population reaches extinction.

FIGURE. Solution curves for harvesting a logistic population with 4h − kM1.