Question

A population $p(t)$ of 1000 bacteria introduced into a nutrient medium grows according to the relation $\text{p}(t) = 1000 + \frac{1000t}{100+t^2}$. The maximum size of this bacterial population is

• A. 1100
• B. 1250
• C. 1050
• D. 950

Hint:

Maximum occurs where derivative = 0.

Answer 1

The Correct Option is C

Step 1: Differentiate $p(t)$. \[ p(t) = 1000 + \frac{1000t}{100+t^2} \] \[ \frac{dp}{dt} = 1000 \cdot \frac{(100+t^2) - 2t^2}{(100+t^2)^2} = 1000 \cdot \frac{100 - t^2}{(100+t^2)^2} \]
Step 2: Set derivative = 0. \[ 100 - t^2 = 0 \Rightarrow t = 10 \]
Step 3: Find maximum value. \[ p(10) = 1000 + \frac{1000 \cdot 10}{100 + 100} = 1000 + \frac{10000}{200} = 1000 + 50 = 1050 \]
Step 4: Conclusion. \[ {1050} \]