If a pond can support up to 10, 000 fish, the rate of growth of the fish population is jointly
proportional to the number of fish present and the difference between 10, 000 and the number of fish
present. The rate of growth is 90 fish per week when 1, 000 fish are present.
a. Find a mathematical model expressing the rate of population growth as a function of the number
present.
b. What is the domain of your function in part (a)?
c. What is the size of the fish population for the growth to be a maximum?
a. To find a mathematical model expressing the rate of population growth as a function of the number present, we can use the concept of joint proportionality. Let's call the rate of growth R and the number of fish present N. According to the problem, we know that the rate of growth is jointly proportional to the number of fish present (N) and the difference between 10,000 and the number of fish present (10,000 - N).
Mathematically, this can be represented as:
R ∝ N * (10,000 - N)
Since we know that the rate of growth is 90 fish per week when 1,000 fish are present, we can substitute these values into the equation:
90 ∝ 1,000 * (10,000 - 1,000)
b. To find the domain of our function in part (a), we need to determine the range of values that are valid for the number of fish present (N). Since a pond can support up to 10,000 fish, the domain of our function will be the set of all values of N that are between 0 and 10,000, inclusive. In interval notation, the domain can be expressed as [0, 10000].
c. To find the size of the fish population for the growth to be a maximum, we need to find the maximum value of the rate of growth function. In the equation R ∝ N * (10,000 - N), the rate of growth (R) will be maximized when the product of N and (10,000 - N) is maximized.
To find the maximum value of the product N * (10,000 - N), we can use calculus. We differentiate the function with respect to N and set the derivative equal to zero to find the critical points, and then determine which of those critical points correspond to a maximum.
Taking the derivative of N * (10,000 - N) with respect to N, we get:
d(N * (10,000 - N))/dN = 10,000 - 2N
Setting this equal to zero and solving for N, we have:
10,000 - 2N = 0
2N = 10,000
N = 5,000
So, the size of the fish population for the growth to be a maximum is 5,000 fish.