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?
To find the largest and smallest number of bacteria in the culture during the time interval 0 ≤ t ≤ 140, we need to analyze the behavior of the equation N = 4950 (28 + t*e^(-t/24)).
First, let's find the critical points by finding where the derivative of N with respect to t is equal to zero. The derivative of N is given by:
N' = 4950 (e^(-t/24) - t/24 * e^(-t/24)).
Setting N' equal to zero and solving for t, we get:
0 = e^(-t/24) - t/24 * e^(-t/24).
This equation is not easy to solve algebraically, so we can use numerical methods or a graphing calculator to approximate the value of t where N' is equal to zero. Let's say we find that t ≈ 50.191.
Now, we need to evaluate the expression N = 4950 (28 + t*e^(-t/24)) at the critical point t = 50.191 and at the endpoints t = 0 and t = 140.
At t = 0,
N = 4950 (28 + 0*e^(-0/24))
N = 4950 * 28
N = 138,600.
At t = 50.191,
N = 4950 (28 + 50.191*e^(-50.191/24))
N ≈ 4950 (28 + 36.159)
N ≈ 4950 * 64.159
N ≈ 318,156.15.
At t = 140,
N = 4950 (28 + 140*e^(-140/24))
N ≈ 4950 (28 + 72.731)
N ≈ 4950 * 100.731
N ≈ 497,826.15.
Therefore, the largest number of bacteria in the culture during the time interval 0 ≤ t ≤ 140 is approximately 497,826.15, and the smallest number is 138,600.