Question

A drug is injected into the body in such a way that the concentration, C, in the blood at time t hours, is given by the function C(t) = 10(e-2t – e-3t). At what time is the concentration of the drug in the body at the highest? Express your

answer to three decimal places.

Answer 1

Answer: time t = 0.405 units

Explanation:

The given function is

C(t) = 10(e-2t – e-3t)

If t = 0, substitute t in the equation

C(0) = 10(e-2(0) – e-3(0))

C(0) = 10 (1 - 1)

C = 0

If t = 1

C(1) = 10(e-2(1) – e-3(1))

C(1) = 10(0.0855)

C(1) = 0.855

1f t = 2

C(2) = 10(e-2(2) – e-3(2))

C(2) = 10( 0.0154)

C(2) = 0.154

If t = 5

C(5) = 10(e-2(5) – e-3(5))

C(5) = 10( 0.0000451)

C(5) = 0.000451

C'(t) = 10 (-2e^-2t + 3e^-3t)

Factorize the equation in the bracket

C'(t) = 10 (e^-2t) (-2 + 3e^-t)

0 = e^-2t

0 = -2 + 3e^-t

2/3 = e^-t

ln 2/3 = -t

-0.405465 = -t

t = 0.405

Therefore the absolute maximum over the used domain occurs at an endpoint (t=1), as indeed the relative maximum occurs outside this interval

Answer 2

0.41 = t

Explanation:

Let C represent the amount of concentration

Let t represent time.

Now, differentiate with respect to t

C'(t) = 10 (-2e^-2t + 3e^-3t)

C'(t) = 10 (e^-2t) (-2 + 3e^-t)

0 = e^-2t ( This is not valid or cannot be considered )

0 = -2 + 3e^-t

2/3 = e^-t

ln 2/3 = t

0.41 = t