Question

Part two of previous question.

Answer 1

Part 1,
http://openstudy.com/study#/updates/5273c73fe4b08ad7c0d9f70a

Answer 2

Part 2:

CONSTANT YIELD HARVESTING
In this problem, we assume that fish are caught at a constant rate \(h\) independent of the size of the fish population, that is, the harvesting rate \(H(y,t)=h\). Then \(y\) satisifies

\(\displaystyle\frac{dy}{dt}=r(1-\frac{y}{K})y-h=f(y)~~~~~~~~~(ii)\)

The assumption of a constant catch rate \(h\) may be reasonable when \(y\) is large but becomes less so when \(y\) is small.

(a) If \(h<rK/4\), show that Eq. (ii) has two equilibrium points \(y_1\) with \(y_1<y_2\); determine these points.

(b) Show that \(y_1\) is unstable and \(y_2\) is asymptotically stable.

(c) From a plot of \(f(y)\) versus \(y\), show that if the initial poplulation \(y_0>y_1\), then \(y\rightarrow y_2\) as \(t\rightarrow\infty\), but if \(y_0<y_1\), then\(y\) decreases as \(t\) increases. Not that \(y=0\) is not an equilbrium point, so if \(y_0<y_1\), then extinction will be reached in a finite time.

(d) If \(h>rK/4\), show that \(y\) decreases to zero as \(t\) increases regardless of the value \(y_0\).

(e) If \(h=rK/4\), whow that there is a single equilibrium point \(y=K/2\) and that this point is semistable. Thus the maximum sustainable yiled is \(h_m=rK/4\), corresponding to the equilibrium value \(y-K/2\). Observe that \(h_m\) has the same value as \(Y_m\) in Problem 1 (d). THe fishery is considered to be overexploited if \(y\) is reduced to a level below \(K/2\).

Answer 3

This stuff is really a lot to take in, no? XD

Answer 4

It is indeed, I will eventually need to fill a tri-fold poster board with this stuff. Make a powerpoint about it, and then give a ten minute speech about all of this.

Answer 5

without having taken a course in diffyQ modeling ... all this is greek to me :/

Answer 6

Oh right, I forgot that you weren't a fan of modeling. Thanks anyway amistre :)