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December 19, 2014

December 19, 2014

Posted by **Rene** on Monday, October 22, 2012 at 3:18pm.

C(t)=50t/t^2+25. Determine the time at which the concentration is highest. Round your answer to the nearest tenth of a minute.

- College Algebra -
**bobpursley**, Monday, October 22, 2012 at 3:49pmC=50t/(t^2+25)

It is a shame you are not in calculus, this is a trivial problem in calculus.

In algebra, this will lead to a quatric equation, not an easy situation. Are you allowed to graph it?

get your calculator, and graph

y=50t/(t^2+25)

I see a max at about 4.8 on this very inaccurate plotter.

http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html

- College Algebra -
**Rene**, Monday, October 22, 2012 at 3:54pmI thought that could be the solution. I graphed it on my calculator and I come up with a maximum of 5. Does this seem correct?

- College Algebra -
**Steve**, Monday, October 22, 2012 at 4:13pmI assume you meant

C(t)=50t/(t^2+25)

Since C(0) = C(∞) = 0, and C(t)>0 for t>0 we know there's a max in there somewhere.

Can't think of any easy algebraic way to find the max, except some trial and error.

C(4) = 4.87

C(5) = 5.00

C(6) = 4.91

5 looks like a good candidate.

C(4.9) = 4.999

C(5.1) = 4.999

Looks like t=5.0 is our guy.

- College Algebra -
**Rene**, Monday, October 22, 2012 at 4:22pmThank you. I looked at the table and 5 was 5 so I thought I was on the right track but just wanted to make sure. Thank you so much!

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