Question

Investigate the following harvesting model both qualitatively and analytically. If a constant number h of fish are harvested from a fishery per unit time, then a model for the population P(t) of the fishery at time t is given by:
dP/dt = P(a − bP) − h, P(0) = P0,
where a, b, h, and P0 are positive constants. Suppose a = 7, b = 1, and h = 49 4 .


Determine whether the population becomes extinct in finite time.

a. The population does not become extinct in finite time.
b. The population becomes extinct in finite time for all values of Po.
c. The population becomes extinct in finite time if Po > 7/2
d. The population becomes extinct in finite time if Po < 7/2
e. The population becomes extinct in finite time if Po= 7/2

If so, find that time. (If not, enter NONE.)
t=_______

Answer 1

Answer:

a. The population does not become extinct in finite time.

Step-by-step explanation:

The model for the population of the fishery is

$dP/dt = P(a-bP)-h, P(0) = P_0$

If we rearrange and replace the constants we have:

$\frac{dP}{P(7-P)-49/4} =dt\\\\-4 (\frac{dP}{4(P-7)P+49}) =dt\\\\-4 \frac{dP}{(2P-7)^2} =dt\\\\-4 \int\frac{dP}{(2P-7)^2} =\int dt\\\\-4(-\frac{1}{2(2P-7)})=t+C\\\\\frac{2}{2P-7}=t+C\\\\ t=0 \,\,\, P(0)=P_0\\\\\frac{2}{2P_0-7}=0+C\\\\C=\frac{2}{2P_0-7}$

Now we can calculate if the population become 0 in any finite time

$\frac{2}{2P-7}=t+\frac{2}{2P_0-7}\\\\\frac{2}{2*0-7}=t+\frac{2}{2P_0-7}\\\\-\frac{2}{7}=t+\frac{2}{2P_0-7}$

To be a finite time, t>0

$t=-\frac{2}{7}-\frac{2}{2P_0-7}=0\\\\-\frac{2}{2P_0-7}=\frac{2}{7}\\\\7-2P_0=7\\\\P_0=0$

We can conclude that the only finite time in which P=0 is when the initial population is 0.

Because P0 is a positive constant, we can say that the population does not become extint in finite time.