Question

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

• A. $1110$
• B. $1250$
• C. $1050$
• D. $5250$

Answer 1

The correct option is C $1050$

$p\left(t\right)=1000+\frac{1000t}{100+t^2}$

$\Rightarrow \frac{dP}{dt}=0$ for maximize.

$\Rightarrow \frac{0+(100+t^2)\cdot 1000-1000t\left(2t\right)}{(100+t^2{)}^2}=0$

$\Rightarrow 1000(100+t^2-2t^2)=0$

$100-t^2=0$

$t=\pm 10$

Maximum will be $t=\pm 10$

$p\left(t\right)=1000+\frac{1000\times 10}{100+100}$

$=1000+\frac{10,000}{200}$

$=1000+50$

$=1050$.