Question

please show all definitions and formulas, thanks!

2. Proportional harvesting is of the form de = -2 (1-1) - he, where h is again a positive constant. This is typical for hunting, where if the population is small, the species will be harder to catch. (a) Show that if h>r, the species always goes extinct. (b) Show that if 0 <h <r, there is a stable equilibrium population, and the species never goes extinct. How does the equilibrium depend on h? (c) Find the maximum value of ha, where x is the equilibrium population you found above. Show that it is the same as what you got for constant rate harvesting. Which method of harvesting is preferable?

Answer 1

2. Consider,

di = rx (1-) – ha

a. Rearranging, we get

$\frac{dx}{dt} = (r-h)x - \frac{rx^2}{L}$

Now, for a population, we are restricted to the domain where

Therefore, when

h>rir-h<0

O< XA0 > . - x(4 – 1) = 2p

And hence, the species will always go extinct.

---------------------

b.

dx = re -mp3-

The fixed points of this equation correspond to

– rz? + Lrt – Lhx = 0 Lh - Orr-L+

Therefore, the roots are

ū-1) 70 = 2

For a stable equilibrium

da = r – 21 – 1 <0

i. For the point to be stable, therefore

(1-4)

7-21(1 - ) -<o o T - 2r + 2h - h h-r<0

Since is a positive constant, therefore,
0 <hr

Thus to have stable equilibrium with a non-zero population, we require

0 <hr

--------

(c) The equilibrium population is

-11=

Therefore,

9(h) = hæ=hl (1 -

To maximize this we set

h=5 1 = 7 10 = (-1) - )= (1)

Therefore the maximum value is

$\frac{r}{2}L\left(1 - \frac{1}{2} \right ) = \frac{rL}{4}$

Answer 2

2. Consider,

di = rx (1-) – ha

a. Rearranging, we get

$\frac{dx}{dt} = (r-h)x - \frac{rx^2}{L}$

Now, for a population, we are restricted to the domain where

Therefore, when

h>rir-h<0

O< XA0 > . - x(4 – 1) = 2p

And hence, the species will always go extinct.

---------------------

b.

dx = re -mp3-

The fixed points of this equation correspond to

– rz? + Lrt – Lhx = 0 Lh - Orr-L+

Therefore, the roots are

ū-1) 70 = 2

For a stable equilibrium

da = r – 21 – 1 <0

i. For the point to be stable, therefore

(1-4)

7-21(1 - ) -<o o T - 2r + 2h - h h-r<0

Since is a positive constant, therefore,
0 <hr

Thus to have stable equilibrium with a non-zero population, we require

0 <hr

--------

(c) The equilibrium population is

-11=

Therefore,

9(h) = hæ=hl (1 -

To maximize this we set

h=5 1 = 7 10 = (-1) - )= (1)

Therefore the maximum value is

$\frac{r}{2}L\left(1 - \frac{1}{2} \right ) = \frac{rL}{4}$