Suppose that the population y of a certain species of fish in a given area of the ocean is
described by the logistic equation
dy/dt = r(1 − y/K)y.
Although it is desirable to utilize this source of food, it is intuitively clear that if too many
fish are caught, then the fish population may be reduced below a useful level and possibly
even driven to extinction. 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 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. To include this effect, the logistic equation is replaced by
dy/dt = r(1 − y/K)y − Ey.
(a) Show that if E < r, then there are two equilibrium points, y1 = 0 and y2 = K(1−E/r) >
0.
(b) Show, by drawing a diagram, that y = y1 is unstable and y = y2 is asymptotically stable.
(c) 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 y2. Find Y as a
function of the effort E; the graph of this function is known as the yield–effort curve.
(d) Determine E so as to maximize Y and thereby find the maximum sustainable yield Ym.

See explanation

Step-by-step explanation:

Given:-

- The schaefer model is given as follows:

                                   dy/dt = r(1 − y/K)y − Ey.

Find:

Show that if E < r, then there are two equilibrium points p1 = 0 and p2 = K · (1 − e/r) > 0.

Solution:

- The equilibrium point is given by setting the ODE dp/dt to zero, we get:

                                P*(r*( 1 - P / k) - E) = 0

                                P1 = 0 or k*( 1 - P / k) - E = 0

                                P1 = 0 or p2 = k · (1 − E/r)

                                P2 > 0 when E > r & r > 0

Find:

- Assume E 0. Also, classify the equilibrium points (source, sink, or node) only for P2.

Solution:

- The stability of P2:

- dp/dt has no further roots in the interval [ 0 ; k · (1 − E/r) ] , Hence, there is no particular sign change in this interval.

- dp/dt > 0 for P = k · (1 − E/r) - E as dp/dt > 0 for p = E

- dp/dt is a quadratic function of p with two roots p1 and p2, Hence, dp/dt changes signs at each root thus dp/dt < 0 for P = k · (1 − E/r) + E.

 - Thus, solutions close to the equilibrium P = (1 − E/r) moves closer to the equilibrium point reaching stability as a node.

Find:

- 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 P2 obtained in part (a) above. Find Y as a function of the effort E.

Solution:

- The sustainable yield (at least if po  > 0 ). is the yield we get in the asymptotically stable equilibrium as all solutions tend to this equilibrium.

                                     y = E . P2 = E.k(1 - E/r)

-  y is a quadratic function of E with roots in 0 and r and a negative second derivative  . The graph is given qualitatively ( Attachment ).

Find:

Determine E so as to maximize Y and thereby find the maximum sustainable yield Ymax.

Solution:

- Maximize y with respect to E, take first derivative of y wrt E:

                              dy / dE = k* ( - 1 / r )*E + k*( 1 - E/r )

                              0 = k*( 1 - 2*E/r )

                               E = r / 2

- As y is a quadratic function with negative second derivative, hence The function y maximizes at E = r / 2. Hence, the maximum sustainable yield y_max is:

                              y_max = y ( r / 2 ) = r*k/ 2 * ( 1 - 1/2)

                              y_max =  r*k/ 2 * ( 1/2)

                              y_max = r*k/ 4

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step-by-step explanation:

i believe it's a.

i'm not sure on this one, most likely correct.