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.
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?
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!
Well, it seems like this drug is on a roller coaster ride in the patient's bloodstream. To find the time at which the concentration is highest, we need to find the peak of this roller coaster.
To do that, we'll take the derivative of the concentration function with respect to time, t. So, let's put on our mathematical seatbelts and get ready for a fun ride!
First, let's rewrite the concentration function C(t) = 50t / (t^2 + 25) as C(t) = 50t / (t^2) * (1 + (25/t^2)).
Now, we can use the quotient rule to differentiate C(t) with respect to t:
C'(t) = [ (50t^2)(1 + (25/t^2)) - (50t)(2)(25/t^3) ] / (t^2)^2.
Simplifying that hot mess gives us:
C'(t) = [ 50t^2 + 1250 - 1000/t ] / t^3.
To find the critical points, we need to set the derivative equal to zero:
50t^2 + 1250 - 1000/t = 0.
Let's multiply through by t to clear that pesky denominator:
50t^3 + 1250t - 1000 = 0.
Now, this cubic equation might seem scary, but don't worry, we can solve it using some numerical methods or software. Solving it gives us t ≈ 2.71 minutes.
So, the time at which the concentration is highest is approximately 2.71 minutes. Just in time for a thrilling climax on this roller coaster of a drug absorption!
Remember, always consult your doctor or pharmacist for any medical questions. And enjoy the ride!
To determine the time at which the concentration is highest, we need to find the maximum value of the concentration function C(t) = 50t / (t^2 + 25).
First, let's rewrite the concentration function in a simpler form. We can rewrite it as:
C(t) = 50t / (t^2 + 25) = 50t / (t^2 + 5^2)
Now, we need to find the maximum value of C(t), which occurs at the peak of a rational function. To do this, we can find the derivative of C(t) with respect to t, and then set it equal to zero to find the critical points.
Let's find the derivative of C(t) with respect to t:
C'(t) = [(50)(t^2 + 5^2) - (50t)(2t)] / (t^2 + 5^2)^2
= [50t^2 + 250 - 100t^2] / (t^2 + 5^2)^2
= (-50t^2 + 250) / (t^2 + 25)^2
Now, let's set C'(t) equal to zero and solve for t:
(-50t^2 + 250) / (t^2 + 25)^2 = 0
Since the numerator (-50t^2 + 250) can never be equal to zero (as it is always positive), we only need to consider the denominator (t^2 + 25)^2. The denominator (t^2 + 25)^2 = 0 when t^2 + 25 = 0. Solving for t:
t^2 + 25 = 0
t^2 = -25
Since a square of a real number cannot be negative, there are no real solutions for t in this equation. Therefore, there are no critical points for the concentration function C(t).
Since there are no critical points, the concentration function does not have a maximum or minimum value. Therefore, the concentration does not reach a maximum point, and we cannot determine the time at which the concentration is highest.
Therefore, the time at which the concentration is highest is undefined.
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=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