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

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(C) Given the population function $p(t) = 1000 + \frac{1000t}{100 + t^2}$.To find the maximum,we differentiate $p(t)$ with respect to $t$:$\frac{dp}{d

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$1050$

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(C) Given the population function $p(t) = 1000 + \frac{1000t}{100 + t^2}$.To find the maximum,we differentiate $p(t)$ with respect to $t$:$\frac{dp}{dt} = 0 + \frac{(100 + t^2)(1000) - (1000t)(2t)}{(100 + t^2)^2}$$\frac{dp}{dt} = \frac{100000 + 1000t^2 - 2000t^2}{(100 + t^2)^2} = \frac{1000(100 - t^2)}{(100 + t^2)^2}$For critical points,set $\frac{dp}{dt} = 0$,which gives $100 - t^2 = 0$,so $t = 10$ (since $t > 0$ for time).For $t  0$ and for $t > 10$,$\frac{dp}{dt} Since the derivative changes from positive to negative at $t = 10$,the function has a local maximum at $t = 10$.The maximum population is $p(10) = 1000 + \frac{1000(10)}{100 + 10^2} = 1000 + \frac{10000}{200} = 1000 + 50 = 1050$.

$A$ 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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