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Mathematics
Problem
A fish population satisfies
d
P
d
t
=
k
P
(
1
−
P
3000
)
\dfrac{dP}{dt}=kP\!\Bigl(1-\dfrac{P}{3000}\Bigr)
d
t
d
P
=
k
P
(
1
−
3000
P
)
, with
P
P
P
measured in fish and
t
t
t
in years. Given
P
(
0
)
=
600
P(0)=600
P
(
0
)
=
600
and
P
(
5
)
=
900
P(5)=900
P
(
5
)
=
900
, determine the value of the constant
k
k
k
(to three decimal places).
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