The concentration of a drug t hours after being injected is given by
C(t) = (0.3t)/(t^2+5). Find the time when the concentration is at a maximum. Give your answer accurate to at least 2 decimal places.

The concentration is at maximum at t = 2.24 hours which is obtained from the first derivative of the function at 0

Given the function :

The maximum value of the function is given by the first derivative of C(t) = 0

Therefore,

dC/dt = [t² - 5(0.3) - 0.3t(2t)] ÷ (t² + 5)²

dC/dt = 0.3t² - 1.5 - 0.6t² / (t² + 5)²

Concentration will be maximum when, dC/dt = 0

Therefore,

0.3t² - 1.5 - 0.6t² / (t² + 5)² = 0

0.3t² - 1.5 - 0.6t² = 0

-0.3t² - 1.5 = 0

-0.3t² = 1.5

t² = (1.5 ÷ 0.3)

t² = 5

t = √5

t = 2.236 hours

Therefore, the concentration will be maximum in 2.24 hours.

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t = 2.236 hours

or 2 hours 14.16 minutes.

Step-by-step explanation:

C = (0.3t)/ (t^2 + 5) where C is the concentration of the drug.

Finding the derivative using the Quotient rule:

C'  =  (t^2 + 5) (0.3) - (0.3t)* 2t /  (t^2 + 5)^2

C' =   0.3t^2 + 1.5 - 0.6t^2 /  (t^2 + 5)^2

C' =  -0.3t^2 + 1.5 / (t^2 + 5)^2

C is at a maximum when the derivative is zero.

Equating this to zero:

 -0.3t^2 + 1.5 / (t^2 + 5)^2 = 0

-0.3t^2 + 1.5  = 0

-0.3t^2 = -1.5

t^2 = -1.5 / -0.3 = 5

t = √5

t = 2.236 hours.

ghjmutmum

step-by-step explanation: