Question

An animal population is modeled by the function A(t) that satisfies the differential equation dA dt equals the product of A divided by 1125 and the quantity 450 minus A . What is the animal population when the population is increasing most rapidly?
45 animals 225 animals 40 animals 180 animals

Answer 1

Answer:

225 animals

Step-by-step explanation:

From the given information:

$\dfrac{dA}{dt} = ( \dfrac{A}{1125}) (450 - A)$

$\dfrac{dA}{dt} = \dfrac{450A - A^2}{1125}$

For a population increasing most rapidly; we have:

$(\dfrac{dA}{dt} \implies max = A')$

Thus; for $\dfrac{d}{dA} ( \dfrac{dA}{dt}) \implies \dfrac{450-2A}{1125}$

$\dfrac{dA'}{dA} \implies 0 \\ \\ \\ A = 225$