A population P-t- of 1000 bacteria introduced into a nutrient medium grows according to P-t- - 1000 - dfrac -1000t-100 - t-2- Then maximum size of bacterial population is1110125010505250

P-t-1000-1000t100-t2-x21D2-dPdt-0 for maximize-x21D2-0-100-t2-x22C5-1000-x2212-1000t-2t-100-t2-2-0-x21D2-1000-100-t2-x2212-2t2-0100-x2212-t2-0t-xB1-10Maximum will be t-xB1-10p-t-1000-1000-xD7-10100-100-1000-10-000200-1000-50-1050-

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Standard XII

Biology

Question

A population $P (t)$ of $1000$ bacteria introduced into a nutrient medium grows according to $P (t) = 1000 + \frac{1000 t}{100 + t^{2}}$ . Then maximum size of bacterial population is
$1110$
$1250$
$1050$
$5250$

1250

5250

1050

1110

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Solution

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$p (t) = 1000 + \frac{1000 t}{100 + t^{2}}$
$\Rightarrow \frac{d P}{d t} = 0$ for maximize.
$\Rightarrow \frac{0 + (100 + t^{2}) \cdot 1000 - 1000 t (2 t)}{(100 + t^{2})^{2}} = 0$
$\Rightarrow 1000 (100 + t^{2} - 2 t^{2}) = 0$
$100 - t^{2} = 0$
$t = \pm 10$
Maximum will be $t = \pm 10$
$p (t) = 1000 + \frac{1000 \times 10}{100 + 100}$
$= 1000 + \frac{10, 000}{200}$
$= 1000 + 50$
$= 1050$ .

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A population $P (t)$ of $1000$ bacteria introduced into a nutrient medium grows according to $P (t) = 1000 + \frac{1000 t}{100 + t^{2}}$ . Then maximum size of bacterial population is
$1110$
$1250$
$1050$
$5250$