Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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

Sagot :

ajeigbeibraheem

Answer:

225 animals

Step-by-step explanation:

From the given information:

[tex]\dfrac{dA}{dt} = ( \dfrac{A}{1125}) (450 - A)[/tex]

[tex]\dfrac{dA}{dt} = \dfrac{450A - A^2}{1125}[/tex]

For a population increasing most rapidly; we have:

[tex](\dfrac{dA}{dt} \implies max = A')[/tex]

Thus; for [tex]\dfrac{d}{dA} ( \dfrac{dA}{dt}) \implies \dfrac{450-2A}{1125}[/tex]

[tex]\dfrac{dA'}{dA} \implies 0 \\ \\ \\ A = 225[/tex]

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.