Step 1: Understanding the Concept:
To find the maximum size of the population, we need to find the global maximum of the function $p(t)$ for $t>0$. This is a standard application of calculus: find the critical points by setting the first derivative to zero, verify it's a maximum, and evaluate the original function at that point.
Step 2: Key Formula or Approach:
1. Differentiate $p(t)$ using the quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.
2. Solve $p'(t) = 0$ to find critical time $t$.
3. Substitute the critical value $t$ back into $p(t)$ to find the maximum population.
Step 3: Detailed Explanation:
The population function is given by:
\[ p(t) = 1000 + \frac{1000t}{100+t^2} \]
Find the derivative $p'(t)$ with respect to $t$. The derivative of the constant 1000 is 0.
Apply the quotient rule to the second term: let $u = 1000t$ and $v = 100+t^2$. Then $u' = 1000$ and $v' = 2t$.
\[ p'(t) = 0 + \frac{(1000)(100+t^2) - (1000t)(2t)}{(100+t^2)^2} \]
Factor out 1000 from the numerator to simplify:
\[ p'(t) = 1000 \left[ \frac{100 + t^2 - 2t^2}{(100+t^2)^2} \right] \]
\[ p'(t) = 1000 \left[ \frac{100 - t^2}{(100+t^2)^2} \right] \]
To find critical points, set $p'(t) = 0$:
\[ 1000 \left[ \frac{100 - t^2}{(100+t^2)^2} \right] = 0 \]
Since a fraction is zero only when its numerator is zero (and denominator is not):
\[ 100 - t^2 = 0 \]
\[ t^2 = 100 \implies t = 10 \text{ or } t = -10 \]
Since time $t$ cannot be negative in this context, we take $t = 10$.
(We can confirm it's a maximum via the first derivative test: for $t<10$, $100-t^2>0$ so $p'(t)$ is positive. For $t>10$, $100-t^2<0$ so $p'(t)$ is negative. The function increases then decreases, confirming a local maximum at $t=10$).
Now, substitute $t = 10$ back into the original population function to find the maximum size:
\[ p(10) = 1000 + \frac{1000(10)}{100 + (10)^2} \]
\[ p(10) = 1000 + \frac{10000}{100 + 100} \]
\[ p(10) = 1000 + \frac{10000}{200} \]
\[ p(10) = 1000 + 50 = 1050 \]
Step 4: Final Answer:
The maximum size of the bacterial population is 1050.