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

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.