#### To determine

**To find:** The value of dPdt that corresponds to a stable population.

#### Answer

The value of dPdt that corresponds to a stable population is 0.

#### Explanation

**Given**:

The given equation for the rate of change of fish population is as below.

dPdt=r0(1−P(t)Pc)P(t)−βP(t) (1)

**Calculation:**

The population P(t) amount remains stable, will indicate that there is no change in the value of population with respect to time. No change indicates that the population amount neither increases nor decreases. The first derivative of population P(t) with respect to time will be zero.

dP(t)dt=0

The value of dPdt that corresponds to a stable population is 0.

#### To determine

**To find:** The stable population level for the pond which can sustain 10,000 fishes, with 4% harvesting rates.

#### Answer

The stable population level for the pond which can sustain 10,000 fishes, with a 5% birth rate, 4% harvesting rate is 2,000.

#### Explanation

**Calculation:**

Substitute 5100 for r0, 4100 for β, 10,000 for Pc in equation (1).

dPdt=5100(1−P(t)10,000)P(t)−4100P(t)

For a stable population condition, substitute 0 for dPdt.

0=5100(1−P(t)10,000)P(t)−4100P(t)4100P(t)=(5100P(t)−5100[P(t)]210,000)0=(1100P(t)−5100[P(t)]210,000)1100P(t)=5100[P(t)]210,0001=5[P(t)]10,000P(t)=2000

The stable population level for the pond which can sustain 10,000 fishes, with a 5% birth rate, 4% harvesting rate is 2,000.

#### To determine

**To find:** The stable population level for the pond which can sustain 10,000 fishes, with a 5 % harvesting rates.

#### Answer

The stable population level for the pond which can sustain 10,000 fishes, with a 5% birth rate, 5% harvesting rate is 0.

#### Explanation

**Calculation:**

Substitute 5100 for r0, 5100 for β, 10,000 for Pc in equation (1).

dPdt=5100(1−P(t)10,000)P(t)−5100P(t)

For a stable population condition, substitute 0 for dPdt.

0=5100(1−P(t)10,000)P(t)−5100P(t)5100P(t)=(5100P(t)−5100[P(t)]210,000)0=(−5100[P(t)]210,000)0=5100[P(t)]210,000P(t)=0

The stable population level for the pond which can sustain 10,000 fishes, with a 5% birth rate, 5% harvesting rate is 0.