Question

A certain population of fish would grow at a relative rate of 20% per month if it had unlimited resources. Unfortunately, the carrying capacity is only 1,000 fish. Fish are harvested at a rate of 48 fish per month. a) Find the two equilibrium solutions. fish (smaller solution) (round to the nearest fish.) fish (larger solution) (round to the nearest fish.) b) There are 300 fish in month t=0. When are there no more fish? After months. (round to the nearest month).

Answer 1

The smaller equilibrium solution is 240 fish and the larger equilibrium solution is 1,000 fish. If there are initially 300 fish, it would take approximately 300 months for the population to reach zero.

Step-by-step explanation:

The equilibrium solutions can be found by setting the growth rate equal to the harvesting rate and solving for the population size. Let's denote the population size as P.

a) Setting the growth rate equal to the harvesting rate: 0.20P = 48

Solving for P: P = 48 / 0.20 = 240

The smaller equilibrium solution is 240 fish. In this case, the population growth rate (0.20 x 240) would just balance out the harvesting rate of 48 fish, resulting in a stable population of 240 fish.

Let's calculate the larger equilibrium solution using the carrying capacity of 1,000 fish:

Setting the growth rate equal to zero (at the carrying capacity): 0.20P - 48 = 0

Solving for P: P = 48 / 0.20 = 240

The larger equilibrium solution is 1,000 fish. In this case, the population growth rate would be zero, as it would just balance out the harvesting rate of 48 fish, resulting in a stable population of 1,000 fish.

b) If there are initially 300 fish, we can calculate how many months it would take for the population to reach zero by setting the growth rate and harvesting rate equal to zero:

0.20P - 48 = 0

Solving for P: P = 48 / 0.20 = 240

Since the smaller equilibrium solution is 240 fish, if there are initially 300 fish, it would take 240 - 300 = -60 fish for the population to reach zero. Therefore, there would be no more fish after negative 60 / 0.20 = -300 months. However, since time cannot be negative, we can round to the nearest month, and the answer would be 300 months.

Answer 2

The equilibrium solutions for the fish population are 240 fish (smaller solution) and 1,000 fish (larger solution). It takes approximately 60 months for the population to decrease to zero starting with 300 fish.

Step-by-step explanation:

To find the equilibrium solutions in this situation, we need to consider the population growth and the rate of harvest. The relative growth rate of the fish population is 20% per month, but the carrying capacity is limited to 1,000 fish. The rate of harvest is 48 fish per month. Equilibrium occurs when the growth rate equals the harvest rate, so we can set up the following equation:

Growth Rate - Harvest Rate = 0

0.20x - 48 = 0

0.20x = 48

x = 48 / 0.20

x = 240

So the smaller equilibrium solution is 240 fish. To find the larger equilibrium solution, we need to consider the carrying capacity. Since the carrying capacity is 1,000 fish, the larger equilibrium solution is the carrying capacity itself, which is 1,000 fish.

b) To find when there are no more fish starting with 300 fish, we need to determine how many months it takes for the population to decrease to zero. We can use the same equation as before, but this time set the growth rate to zero:

0.20x - 48 = 0

0.20x = 48

x = 48 / 0.20

x = 240

Since we start with a population of 300 fish, we need to subtract 240 from 300 to determine how many months it takes for the population to decrease to zero:

300 - 240 = 60

So it takes approximately 60 months for the population to decrease to zero.