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March 31, 2015

March 31, 2015

Posted by **Elisa** on Monday, November 29, 2010 at 9:52pm.

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

- Calculus -
**Reiny**, Monday, November 29, 2010 at 10:27pmdN/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?

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