Given :
The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by [tex]K(x) = \dfrac{5x}{x^2+9}[/tex].
To Find :
Find the time at which the concentration is a maximum. b. Find the maximum concentration.
Solution :
For maximum value of x, K'(x) = 0.
[tex]K'(x) = \dfrac{5(x^2+9)- 5x(2x)}{(x^2+9)^2}=0\\\\5x^2+45-10x^2=0\\\\5x^2 = 45\\\\x = \pm 3[/tex]
Since, time cannot be negative, so ignoring x = -3 .
Putting value of x = 3, we get, K(3) = 15/( 9 + 9) = 5/6
Therefore, maximum value drug in bloodstream is 5/6 at time x = 3 units.
Hence, this is the required solution.
