A) 1100
B) 1250
C) 1050
D) 5250
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 \[{{\left. \frac{dp}{dt} \right|}_{t\,<\,10}}>0\] and \[{{\left. \frac{dp}{dt} \right|}_{t\,>\,10}}<0\] \ At \[t=10\], \[\frac{dp}{dt}\]change from positive to negative. \ p is maximum at \[t=10\]. \ \[{{p}_{\max }}=p(10)=1000+\frac{1000.10}{100+{{10}^{2}}}=1050\].
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