OPTIMIZATION

Question

THE CONCENTRATION C(t) in milligrams pee cubic centimeter of a particular drug in a patients bloodstream is given by: C(t)= 0.16t/t^2+4t+4 where t is the number of hours after the drug is taken. how many hours after the drug is given will the concentration be maximum? What is the max concentration?

Answer 1

Set the derivative of C(t) equal to zero.

 

d/dt[C(t)] = 0

 

 

Solve for x.  This is your critical value.  Evaluate the derivative before and after this critical value.

 

If the derivative changes from positive to negative, then you have a maximum at the critical value.

If the derivative changes from negative to positive, then you have a minimum at the critical value.

 

Note.  Your vertical asymptote is x=-2

 

 

d/dx  [0.16t / (t + 2)²] = 0

 

[0.16(t + 2)² - 2(0.16t)(t + 2)] / (t + 2)⁴ = 0

 

0.16(t + 2)^-2 - 0.32t(t + 2)^-3 = 0

 

(t + 2)^-3[0.16(t + 2) - 0.32t] = 0

 

(t + 2)^-3(0.32 - 0.16t) = 0

 

 

0.32 - 0.16t = 0

       - 0.16t = -0.32

                t = 2

 

t = 2    --->  critical value