Asked by Rene
The concentration C of a certain drug in a patient's bloodstream t minutes after injection is given by
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.
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.
Answers
Answered by
bobpursley
C=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
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
Answered by
Rene
I 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?
Answered by
Steve
I 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.
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.
Answered by
Rene
Thank 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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