Question

An animal population is modeled by the function A(t) that satisfies the differential equation dA/dt= A/1125 all this multiplied by (450-A). What is the animal population when the population is increasing most rapidly?
a) 45 animals
b) 225 animals
c) 40 animals
d) 180 animals

Answer 1

all this ????
dA/dt= (A/1125 )(450-A) ???? maybe
dA/dt = 0.4 A - A^2 /1125

well if y = 0.4 x - .00035555 x^2
when is y max?
at the vertex where dy/dx = 0
0 = .4 - .00071111 x
x = .4/.00071111 = 562
I guess my guess at what your question means is incorrect.

Answer 2

An animal population is modeled by the function A(t) that satisfies the differential equation dA/dt= A/1125 and the quantity (450-A). What is the animal population when the population is increasing most rapidly?
I have only these options (on the top)
a)45
b)225
c)40
d)180
Please can you review again this question?
Thanks in advance

Answer 3

I am sorry I do not know what this means:

dA/dt= A/1125 and the quantity (450-A)

Answer 4

me neither, but thanks anyway.