Silly math!!?

Question

1 The number of bacteria in a culture at time t is given approximately by y=1000(25 + t e^(t/20)) for 0<=t <=100.
(a) Find the largest number and the smallest number of bacteria in the culture during the interval.
(b) At what time during the interval is the rate of change in the number of bacteria minimum?

Answer 1

y=1000(25 + t e^(t/20)) for 0<=t <=100.
y = 25000 + 1000te^(t/20)
y - 25000 = 1000te^(t/20)
ln(y-2500) = ln[1000te^(t/20)]
ln(y-2500) = (t/20)ln(1000te)
y'/(y-2500) =
you get the point
take the derative and you should get the answer
i would do it but its vacation

Answer 2

smallest number when t =0 = 25,000
largest number when t = 100 = 14866315.91