Calculus
posted by Elisa on .
Suppose that the number of bacteria in a culture at time t is given by
N = 4950 ( 28 + t*e ^(t/24))
Find the LARGEST and SMALLEST number of bacteria in the culture during the time interval
0 ≤ t ≤ 140

dN/dt = 4950( 0 + (1/24)t(e^(t/24) + e^(t/24)
= 0 for a max/min
...
...
(1/24)t (e^(t/24)) = e^(t/24)
(1/24)t = 1
t = 24
so t=24 gives a local max/min
but we also have to consider the endpoints.
f(0) = 4950(28 + 0) = 138 600
f(24) = 4950(28 + 8.8291) = 182 304
f(140) = 4950(28 + .40996) = 140 629
So what do you think?