A population p(t) of 1000 bacteria introduced | vedclass

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(C) To find the maximum size of the population,we need to find the maximum value of the function $p(t) = 1000 + \frac{1000t}{100 + t^2}$.Let $f(t) = \

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$1050$

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(C) To find the maximum size of the population,we need to find the maximum value of the function $p(t) = 1000 + \frac{1000t}{100 + t^2}$.Let $f(t) = \frac{1000t}{100 + t^2}$. To find the critical points,we calculate the derivative $f'(t)$ and set it to $0$.Using the quotient rule,$f'(t) = 1000 	imes \frac{(100 + t^2)(1) - t(2t)}{(100 + t^2)^2} = 1000 	imes \frac{100 - t^2}{(100 + t^2)^2}$.Setting $f'(t) = 0$ gives $100 - t^2 = 0$,so $t^2 = 100$,which implies $t = 10$ (since $t \ge 0$).Now,we evaluate $p(t)$ at $t = 10$:$p(10) = 1000 + \frac{1000(10)}{100 + (10)^2} = 1000 + \frac{10000}{100 + 100} = 1000 + \frac{10000}{200} = 1000 + 50 = 1050$.Thus,the maximum size of the bacterial population is $1050$.

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

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