Applications Of Derivatives Question 183
Question: A population p(t) of 1000 bacteria introduced into nutrient medium grows according to the relation $ p(t)=1000+\frac{1000t}{100+t^{2}} $ . The maximum size of this bacterial population is
[Karnataka CET 2005]
Options:
A) 1100
B) 1250
C) 1050
D) 5250
Show Answer
Answer:
Correct Answer: C
Solution:
$ p(t)=1000+\frac{1000t}{100+t^{2}} $
$ \frac{dp}{dt}=\frac{(100+t^{2}),1000-1000t,.,2t}{{{(100+t^{2})}^{2}}} $
$ =\frac{1000,(100-t^{2})}{{{(100+t^{2})}^{2}}} $
For extremum, $ \frac{dp}{dt}=0\Rightarrow t=10 $
Now $ {{. \frac{dp}{dt} |} _{t,<,10}}>0 $ and $ {{. \frac{dp}{dt} |} _{t,>,10}}<0 $
At $ t=10 $ , $ \frac{dp}{dt} $ change from positive to negative.
p is maximum at $ t=10 $ .
$ {p _{\max }}=p(10)=1000+\frac{1000.10}{100+10^{2}}=1050 $ .
BETA
