Constant Effort Harvesting. At a given level of effort, it is reasonable to assu

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Constant Effort Harvesting. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population y: the more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by H(y, t) = Ey, where E is a positive constant, with units of 1/time, that measures the total effort made to harvest the given species of fish. With this choice for H(y, t). Eq. (1) becomes dy/dt = r(1 - y/K)y- Ey. This equation is known as the Schaefer model after the biologist M. B. Schaefer, who applied it to fish populations. Show that if E 0. Show that y = y1 is unstable and y1 = y2 is asymptotically stable. A sustainable yield Y of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort E and the asymptotically stable population v'2- Find Y as a function of the effort E. The graph of this function is known as the yield-effort curve. Determine E so as to maximize Y and thereby find the maximum sustainable yield Ym. 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 satisfies dy/dt = r( 1 -y/K)y-h = f(y). The assumption of a constant catch rate h may be reasonable when>> is large but becomes less so when y is small. If h

Explanation / Answer

The equilibrium points of an autonomous differential equation are the values of the dependent variable, y, for which dy/dt = 0

In this case, we have that

dy/dt = r(1 - y/K)y - Ey

Set dy/dt equal to zero and solve for y to determine the equilibrium points:

0 = ry - r*K*y^2 - Ey

0 = (r - E)*y - r*y^2/K

0 = y*[(r-E) - (r/K)*y]

y_eq1 = 0 and y_eq2 = (r-E)/(r/K) = K(1 - E/r)

Note that these are always the equilibrium points of this equation; however, the second point is only physically reasonable if E < r (as long as E< r, y_eq2 > 0. If r < E, y_eq2 is negative, which is not a physically reasonable value for a population.)

b) If d^2y/dt^2 < 0 at an equilibrium point, then the equilibrium is asymptotically stable. If the second derivative is positive then the equilibrium is unstable. Let's see what the story is in this case.

y'' = r - E - 2*r*y/K

at y = 0, y'' = r - E, but we are told that r > E, so r - E > 0 and this is an unstable equilibrium.

at y = K(1 - E/r), y'' = r - E - 2r - 2E = -r - 3E.

We are told that E is a positive constant, and r > E, so r must also be positive, so -r - 3E is negative, and the second equilibrium point is stable.

c) We have that Y = E*y_eq2 = E*K*(1 - E/r) = K*E - K*E^2/R

You can use this to make a plot of K vs E. It should be a parabola open downward.

d) To find the value of E that maximizes Y, take the derivative of Y with respect to E, set that result equal to zero, and solve for E.

Y = K*E - K*E^2/R

dY/dE = K - 2*K*E/R

0 = K - 2*K*E_max/R

2*K*E_max = K*R

E_max = R/2

Plug this back into the expression for Y as a function of E to determine Y_max:

Y_max = K*R/2 - K*(R/2)^2/R

Y_max = K*R/2 - K*R/4 = K*R/4

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