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 find the maximum and minimum values of the function N(t) = 4950(28 + te^(-t/24)).
To find the maximum and minimum values of a function, we can use calculus. Specifically, we need to find the points where the derivative of the function is equal to zero, and then check the behavior of the function around those points.
First, let's find the derivative of N(t) with respect to t. To do this, we can use the product rule and the chain rule:
N'(t) = 4950(e^(-t/24))(28) + 4950(te^(-t/24))(-1/24)
Simplifying this equation, we have:
N'(t) = 4950e^(-t/24)(28 - t/24)
Now, let's set N'(t) equal to zero and solve for t:
4950e^(-t/24)(28 - t/24) = 0
To find the values of t that satisfy this equation, either the factor on the left side must be zero or the factor on the right side must be zero. Setting them equal to zero separately, we have:
e^(-t/24) = 0 (which has no solution)
or
28 - t/24 = 0
Solving for t in the second equation, we get:
t = 24 * 28
t = 672
So, t = 672 is the only value that satisfies N'(t) = 0.
Now, let's check the behavior of the function N(t) around t = 672 to determine if it gives us the maximum or minimum number of bacteria.
To do this, we can take the second derivative of N(t) with respect to t:
N''(t) = 4950((-1/24)e^(-t/24))(28 - t/12) + 4950(e^(-t/24))(-1/24)
Simplifying this equation, we have:
N''(t) = 4950e^(-t/24)(-1/24)(28 - t/12 - 1/24)
N''(t) = 4950e^(-t/24)(-1/24)(-t/12 - 95/24)
N''(t) = 4950e^(-t/24)((t/12) + 95/24) / 24
To determine if the function has a maximum or minimum at t = 672, we need to substitute this value into the second derivative:
N''(672) = 4950e^(-672/24)((672/12) + 95/24) / 24
Simplifying this equation, we have:
N''(672) = 4950e^(-672/24)(56 + 95/24) / 24
N''(672) = 4950e^(-672/24)(1343/24) / 24
Next, we need to evaluate the sign of N''(672) to determine if it is positive or negative. If N''(672) is positive, then the function has a minimum at t = 672. If N''(672) is negative, then the function has a maximum at t = 672.
Calculating the value of N''(672), we have:
N''(672) = 4950e^(-672/24)(1343/24) / 24
N''(672) ≈ -2127.56
Since N''(672) is negative, it means the function has a maximum at t = 672.
Therefore, to find the largest number of bacteria in the culture during the time interval 0 ≤ t ≤ 140, we substitute t = 672 into the function N(t):
N(672) = 4950(28 + (672)e^(-672/24))
Calculating this value, we get:
N(672) ≈ 1.18
So, the largest number of bacteria in the culture during the time interval 0 ≤ t ≤ 140 is approximately 1.18.
To find the smallest number of bacteria, we need to evaluate N(0) and N(140) and compare them.
N(0) = 4950(28 + (0)e^(-0/24))
N(0) = 4950(28 + 0)
N(0) = 4950(28) = 138,600
N(140) = 4950(28 + (140)e^(-140/24))
Evaluating this equation, we get:
N(140) ≈ 4,570,087
So, the smallest number of bacteria in the culture during the time interval 0 ≤ t ≤ 140 is approximately 138,600.