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During an experiment, the growth rate of a bacteria colony

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During an experiment, the growth rate of a bacteria colony  [#permalink]

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New post Updated on: 27 Aug 2014, 05:02
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During an experiment, the growth rate of a bacteria colony is recorded. It is found that, to a good approximation, the growth rate of the number of bacteria in the colony is given by

\(R=\sqrt{N}(2000-(20-T)^2)\),

where R is the growth rate in millions per hour, N is the number of bacteria, in millions, at the beginning of the experiment, and T is the time, in minutes, that had passed since the beginning of the experiment. The experiment is 1 hour long, so T varies between zero and 60. What is the maximum growth rate of the bacteria colony recorded during the experiment?

A. 400√N millions of bacteria per hour
B. √N millions of bacteria per hour
C. 2000 millions of bacteria per hour
D. 2000√N millions of bacteria per hour
E. 1600 millions of bacteria per hour

Originally posted by goodyear2013 on 27 Aug 2014, 04:34.
Last edited by Bunuel on 27 Aug 2014, 05:02, edited 1 time in total.
Edited the question.
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Re: During an experiment, the growth rate of a bacteria colony  [#permalink]

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New post 27 Aug 2014, 05:06
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goodyear2013 wrote:
During an experiment, the growth rate of a bacteria colony is recorded. It is found that, to a good approximation, the growth rate of the number of bacteria in the colony is given by

\(R=\sqrt{N}(2000-(20-T)^2)\),

where R is the growth rate in millions per hour, N is the number of bacteria, in millions, at the beginning of the experiment, and T is the time, in minutes, that had passed since the beginning of the experiment. The experiment is 1 hour long, so T varies between zero and 60. What is the maximum growth rate of the bacteria colony recorded during the experiment?

A. 400√N millions of bacteria per hour
B. √N millions of bacteria per hour
C. 2000 millions of bacteria per hour
D. 2000√N millions of bacteria per hour
E. 1600 millions of bacteria per hour


Given that \(R=\sqrt{N}(2000-(20-T)^2)\).

To maximize R we need to maximize (2000-(20-T)^2).
To maximize (2000-(20-T)^2) we need to minimize (20-T)^2 (the value subtracted from 2000).
The minimum value of (20-T)^2 is 0 for T = 20.

Therefore, the maximum value of R is \(R=\sqrt{N}(2000-0)=2000\sqrt{N}\).

Answer: D.

Hope it's clear.
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Re: During an experiment, the growth rate of a bacteria colony  [#permalink]

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New post 03 Oct 2018, 10:23
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Re: During an experiment, the growth rate of a bacteria colony   [#permalink] 03 Oct 2018, 10:23
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