Page 183 - How To Solve Word Problems In Calculus

5000
P (t) =
1 + 4e −0.4t
)
= 5000(1 + 4e −0.4t −1
) (−1.6e

P (t) =−5000(1 + 4e −0.4t −2 −0.4t )
8000e −0.4t
=
)
(1 + 4e −0.4t 2
8000e −2

The growth rate after 5 months is P (5) = ≈ 456
(1 + 4e )
−2 2
fish per month.

(d) The population growth rate will begin to decline when P (t)
turns from positive to negative. Since P (t) is a continuous

function, we must determine where P (t) = 0.
8000e −0.4t

P (t) =
(1 + 4e −0.4t 2
)
d d
)
(1 + 4e −0.4t 2 (8000e −0.4t ) − (8000e −0.4t ) (1 + 4e −0.4t 2
)
dt dt

P (t) =
)
(1 + 4e −0.4t 4
(1 + 4e −0.4t 2 −0.4t ) − (8000e −0.4t )(2)(1 + 4e −0.4t )(−1.6e −0.4t )
) (−3200e
=
)
(1 + 4e −0.4t 4
−3200e −0.4t (1 + 4e −0.4t )[(1 + 4e −0.4t ) − 8e −0.4t ]
=
(1 + 4e −0.4t 4
)
−3200e −0.4t [(1 + 4e −0.4t ) − 8e −0.4t ]
=
)
(1 + 4e −0.4t 3
−3200e −0.4t [1 − 4e −0.4t ]
0 =
)
(1 + 4e −0.4t 3

The only way P (t)can be0isif1 − 4e −0.4t = 0.
1 − 4e −0.4t = 0
1 = 4e −0.4t
1
e −0.4t =
4
170