(Resubmission by request) need help with only part D of this problem. The previous parts should help figure out that part. Thanks! (a) Find the equilibrium population as a function of h. What is the largest h consis-N* = 1.5 N - N* )-h N*N-1.SN* (1-N*)+hN* = 002012. Consider the harvest population modelN41 =1.5N (1 Nt) - hN.a the equilibrium population as a function of h. What is the largest h consis-tent with a nonnegative equilibrium?Tet N be equilibriumON-1.SN CI-N*)+hN* = 0N* [I-1 S-N*)+n] =0or N*LI-1SC1-N*n] =0%3D(b) Find the equilibrium harvest as a function of h.P(n)=hN*Ith%3D=h (1-)1.51.5(c) Find the harvesting effort that maximizes the harvest.%3D1.5PO S) = 0.S(1-)(d) Find the equilibria for this harvesting effort. For each, determine if stable orunstable.

Name: (Resubmission by request) need help with only part D of this problem.
Uploaded: 2024-03-05T11:52:21.000Z
Channel: Bartleby
Description: (Resubmission by request) need help with only part D of this problem. The previous parts should help figure out that part. Thanks! (a) Find the equilibrium population as a function of h. What is the largest h consis-N* = 1.5 N - N* )-h N*N-1.SN* (1-N

Given the harvest population model Nt+1=1.5Nt1-Nt-hNt ---(1). Note: We have maximized the function P(h)=h-h+h21.5 written by you in part (b) to obtain the optimal harvesting effort h=0.25 which is the answer for part (c). (d) So substituting h=0.25 in (1), we get Nt+1=1.5Nt1-Nt-0.25NtLet N* be the equilibria, thenN*=1.5N*1-N*-0.25N*⇒N*=0 or 1-1.51-N*+0.25=0 N*=16. Hence, N*=0, 16 are the equilibria for the harvesting effort h=0.25 For stability, we write the given equation in the form Nt+1=FNt. Nt+1=1.5Nt1-Nt-0.25NtF(N)=1.5N(1-N)-0.25N=1.25N-1.5N2.F'(N)=1.25-3NNow, F'(0)=1.25⇒F'(0)>1Hence we conclude that N*=0 is an unstable equilibria.And, F'16=34⇒F'16<1Hence, we conclude that N*=16 is a stable equilibra. Conclusion: The equilibrium points for h=0.25 are 0 and 16. The equilibrium point 0 is unstable while the equilibrium point 16 is stable.