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 .
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 given equation and its corresponding exponential term.
The equation representing the number of bacteria in the culture at time t is:
N = 4950 * (28 + t * e^(-t/24))
To find the largest and smallest numbers, we can focus on the exponential term e^(-t/24) since the rest of the equation is constant.
1. To find the largest number of bacteria:
Let's analyze the exponential term. As t approaches infinity, e^(-t/24) will approach 0. This means that the exponential term will have its smallest positive value (almost 0) which would allow the overall equation to reach its largest value.
Since we have the time interval 0 ≤ t ≤ 140 and we want to find the largest value, we are interested in the value at t = 140.
Plug in t = 140 into the exponential term:
e^(-140/24) ≈ 3.075e-09
Now, plug this value back into the original equation:
N = 4950 * (28 + 140 * 3.075e-09)
= 4950 * (28 + 4.305e-07)
= 137,705.4 (approximately)
Therefore, the largest number of bacteria in the culture during the time interval 0 ≤ t ≤ 140 is approximately 137,705.4.
2. To find the smallest number of bacteria:
In this case, we need to find the smallest value for the exponential term, which occurs when t = 0.
Plug in t = 0 into the exponential term:
e^(-0/24) = e^0 = 1
Now, plug this value back into the original equation:
N = 4950 * (28 + 0)
= 4950 * 28
= 138,600
Therefore, the smallest number of bacteria in the culture during the time interval 0 ≤ t ≤ 140 is 138,600.
In summary, the largest number of bacteria is approximately 137,705.4, and the smallest number is 138,600 within the time interval 0 ≤ t ≤ 140.
To find the largest and smallest number of bacteria in the given time interval, we need to find the maximum and minimum values of the function N(t) = 4950(28 + t*e^(-t/24)) within the range 0 ≤ t ≤ 140.
To find the maximum and minimum values of a function, we can differentiate it and set the derivative equal to zero. Let's find the derivative of N(t) with respect to t:
N'(t) = 4950 * [(0) + (1)(e^(-t/24)) + t * (e^(-t/24)) * (-1/24)]
= 4950 * (e^(-t/24)) * (1 - t/24)
Now, to find the critical points, we set N'(t) = 0 and solve for t:
e^(-t/24) * (1 - t/24) = 0
Since e^(-t/24) is never zero, we have:
1 - t/24 = 0
t/24 = 1
t = 24
So, the critical point is t = 24.
Next, we need to check the endpoints of the interval, t = 0 and t = 140.
Let's evaluate N(t) at these points:
N(0) = 4950(28 + (0)e^(-0/24)) = 4950(28) = 138,600
N(140) = 4950(28 + (140)e^(-140/24)) = 4950(28 + 140*exp(-5.833)) ≈ 159,240 (rounded to the nearest whole number)
To determine whether the critical point t = 24 is a maximum or minimum, we can use the second derivative test. Let's find the second derivative N''(t):
N''(t) = 4950 * [-(e^(-t/24)) * (-1/24) + (t*e^(-t/24)) * (-1/24^2)]
= 4950 * (e^(-t/24)) * (1/24 - t/24^2)
Now, let's evaluate N''(24):
N''(24) = 4950 * (e^(-24/24)) * (1/24 - 24/24^2) = 4950 * (e^(-1)) * (1/24 - 1/24) = 0
Since the second derivative is zero, the second derivative test is inconclusive. However, we can observe that at t = 24, the behavior of the function changes from increasing to decreasing. Thus, we can conclude that t = 24 is the point of local maximum.
Therefore, within the interval 0 ≤ t ≤ 140, the largest number of bacteria in the culture is approximately 159,240, and the smallest number of bacteria is 138,600.