Question

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

• A. $1100$
• B. $1250$
• C. $1050$
• D. $5250$

Answer 1

The correct option is C

$1050$

Explanation for the correct option:

Find the maximum size of bacterial population:

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

For maximization

$\frac{dp}{dt}=0$

$\Rightarrow$$\frac{[0+(100+t^2)\times 1000–1000t(2t\left)\right]}{{[100+t^2]}^2}=0$

$\Rightarrow$ $1000\left(100+t^2–2t^2\right)=0$

$\Rightarrow$ $100–t^2=0$

$\Rightarrow$ $t=\pm 10$

Now, The maximum will be at $t=\pm 10$

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

$$\begin{gathered} =1000+\frac{10000}{200} \\ =1000+50 \\ =1050 \end{gathered}$$

$\therefore$The maximum size of bacterial population is $\mathbf{1050}$

Hence, Option ‘C’ is Correct.