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Question 1186153: In a limited environment where A is the maximum number of bacteria supportable by the environment, the rate of bacterial growth is directly proportional to the number present and the difference between A and the number present. Suppose 1 million bacteria is the maximum number supportable by the environment and the rate of growth is 60 bacteria per minute when there are 1000 bacteria present.
Find the rate of growth when there are 100,000 bacteria present.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**1. Set up the differential equation:**
Let *B(t)* be the number of bacteria at time *t*. The problem states that the rate of growth is directly proportional to the number present and the difference between A and the number present. This can be written as:
dB/dt = k * B * (A - B)
where *k* is the proportionality constant.
**2. Use the given information to find *k*:**
We know that A = 1,000,000, and when B = 1000, dB/dt = 60. Plugging these values into the equation:
60 = k * 1000 * (1,000,000 - 1000)
60 = k * 1000 * 999,000
k = 60 / (1000 * 999,000)
k = 6 / 999,000
k ≈ 6.006 x 10⁻⁸
**3. Find the rate of growth when B = 100,000:**
Now we want to find dB/dt when B = 100,000. Using the same equation and the calculated value of *k*:
dB/dt = (6.006 x 10⁻⁸) * 100,000 * (1,000,000 - 100,000)
dB/dt = (6.006 x 10⁻⁸) * 100,000 * 900,000
dB/dt = 5405.4 bacteria per minute
**Answer:**
The rate of growth when there are 100,000 bacteria present is approximately 5405.4 bacteria per minute.
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