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
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
I am sorry I do not know what this means:
dA/dt= A/1125 and the quantity (450-A)
me neither, but thanks anyway.
To find the animal population when it is increasing most rapidly, we need to find the value of A at which the derivative, dA/dt, is maximum.
Given the differential equation dA/dt = A/1125 * (450-A), we can solve for dA/dt = 0 to find the critical points where the population is either increasing rapidly or decreasing rapidly.
dA/dt = 0
A/1125 * (450-A) = 0
Now, there are two possibilities for dA/dt to be zero: either A = 0 or A = 450.
If A = 0, then dA/dt would be 0, meaning the population is not changing. Therefore, this is not the point where the population is increasing most rapidly.
Now, let's consider A = 450. Substituting A = 450 into the differential equation:
dA/dt = A/1125 * (450-A)
dA/dt = 450/1125 * (450-450)
dA/dt = 0
In this case, dA/dt is also 0, meaning the population is not changing. Therefore, this is not the point where the population is increasing most rapidly.
Hence, neither A = 0 nor A = 450 gives us the maximum value of dA/dt.
To find the maximum point, we can use the first derivative test by taking the second derivative, d^2A/dt^2, and checking its sign.
Taking the second derivative of the original differential equation:
d^2A/dt^2 = d/dt (dA/dt)
d^2A/dt^2 = d/dt (A/1125 * (450-A))
d^2A/dt^2 = (dA/dt * (450-A) + A/1125 * (-1))
Now, plug in A = 225 into the second derivative:
d^2A/dt^2 = (dA/dt * (450-225) + 225/1125 * (-1))
d^2A/dt^2 = (dA/dt * 225 + (-1/5))
Let's take dA/dt = 0 and substitute it into the equation:
d^2A/dt^2 = (0 * 225 + (-1/5))
d^2A/dt^2 = (-1/5)
Since d^2A/dt^2 is negative at A = 225, this means that at A = 225, the derivative dA/dt is maximum, and the population is increasing most rapidly.
Therefore, the answer is b) 225 animals.
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.