A population `p(t)` of 1000 bacteria inroduced into nutrient medium grows according to the relation `p(t)=1000+(1000t)/(100+t^2)`. The maximum size of this bacterial population is

Options

A. 1100
B. 1250
C. 1050
D. 5250

Explanation

To find the maximum size of the bacterial population given by the function \( p(t) = 1000 + \frac{1000t}{100 + t^2} \), we will follow these steps:

### Step 1: Differentiate the population function
We need to find the derivative of \( p(t) \) with respect to \( t \):
\[ p(t) = 1000 + \frac{1000t}{100 + t^2} \]
Using the quotient rule for differentiation, where if \( u = 1000t \) and \( v = 100 + t^2 \), we have:
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
Calculating \( \frac{du}{dt} = 1000 \) and \( \frac{dv}{dt} = 2t \), we can substitute these into the quotient rule:
\[ \frac{dp}{dt} = \frac{(100 + t^2)(1000) - (1000t)(2t)}{(100 + t^2)^2} \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( \frac{dp}{dt} = 0 \):
\[ (100 + t^2)(1000) - (1000t)(2t) = 0 \]
This simplifies to:
\[ 1000(100 + t^2) - 2000t^2 = 0 \]
\[ 100000 + 1000t^2 - 2000t^2 = 0 \]
\[ 100000 - 1000t^2 = 0 \]
\[ 1000t^2 = 100000 \]
\[ t^2 = 100 \quad \Rightarrow \quad t = 10 \text{ (since time cannot be negative)} \]

### Step 3: Verify if it is a maximum
To confirm that \( t = 10 \) gives a maximum, we can check the sign of \( \frac{dp}{dt} \) around \( t = 10 \):
- For \( t < 10 \), \( \frac{dp}{dt} > 0 \) (population is increasing).
- For \( t > 10 \), \( \frac{dp}{dt} < 0 \) (population is decreasing).
Thus, \( t = 10 \) is indeed a maximum.

### Step 4: Calculate the maximum population
Now, we substitute \( t = 10 \) back into the original population function:
\[ p(10) = 1000 + \frac{1000 \cdot 10}{100 + 10^2} \]
Calculating this gives:
\[ p(10) = 1000 + \frac{10000}{100 + 100} = 1000 + \frac{10000}{200} = 1000 + 50 = 1050 \]

### Conclusion
The maximum size of the bacterial population is \( \boxed{1050} \).