Example 4 dealt with the case 4h kM 2 in the equation dx dt kx (M x) h that describes constant rate harvesting of a logistic population Problems 26 and 27 deal with the other cases If 4h kM 2 , show that typical solution curves look as illustrated in Fig 2 2 14 Thus if x 0 M 2, then x (t) M 2 as t But if x 0 M 2, 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 X x 0 t x M 2 FIGURE 2 2 14 Solution curves for harvesting a logistic population with 4h KM

Question: Example 4 dealt with the case 4h > kM 2 in the equation dx/dt = kx (M - x) - h that describes constant-rate harvesting

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 26 and 27 deal with the other cases.

If 4h = kM², show that typical solution curves look as illustrated in Fig. 2.2.14. Thus if x₀ ≧ M/2, then x (t) → M/2 as t →+ ∞. But if x₀

X x=0 t x = M/2 FIGURE 2.2.14. Solution curves for harvesting

a logistic population with 4h = KM.

X x=0 t x = M/2 FIGURE 2.2.14. Solution curves for harvesting a logistic population with 4h = KM.

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