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Calculus

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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

  • Calculus -

    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?

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