Asked by Meredith
                Can we try this again? I already posted this question and thank you for responding but I just don't understand. I am posting it again because I need to clarify that I have formulas my professor has us use but since I've never come across a problem like this one before I am just a bit confused.
Quadratic Function problem:
When a certain drug is taken orally, the concentration of the drug in the patients bloodstream after t minutes is given by C(t)=0.06t-0.0002t^2, where 0 ≤ t ≤ 240 and the concentration is measured by mg/L. When is the maximum serum concentration reached, and what is that maximum concentration?
    
In class we were given formulas such as:
f(x)= ax^2 + bx+ c
x= -b/2a
And then an f(x) or whatever the letters are being used in the word problem where you plug in the answer for x back into the original equation. I just don't understand what role the 0 ≤ t ≤ 240 comes in because we haven't done any like this in class yet but it was assigned for hw! help?
            
        Quadratic Function problem:
When a certain drug is taken orally, the concentration of the drug in the patients bloodstream after t minutes is given by C(t)=0.06t-0.0002t^2, where 0 ≤ t ≤ 240 and the concentration is measured by mg/L. When is the maximum serum concentration reached, and what is that maximum concentration?
In class we were given formulas such as:
f(x)= ax^2 + bx+ c
x= -b/2a
And then an f(x) or whatever the letters are being used in the word problem where you plug in the answer for x back into the original equation. I just don't understand what role the 0 ≤ t ≤ 240 comes in because we haven't done any like this in class yet but it was assigned for hw! help?
Answers
                    Answered by
            Reiny
            
    C(t)=0.06t-0.0002t^2 
is a quadratic function of the form you were given
f(x) = ax^2 + bx + c
a = -.0002
b = .06
c = 0
x = -b/(2a) gives you the x value where the max value of your function exists, so
t = -.06/(2(-.0002)) = 150
now sub that back into the equation to find the actual concentration
C(150) = .06(150) - .0002(150)^2 = 4.5 units
notice that t = 150 falls within the domain given of 0 ≤ t ≤ 240 , the reason probably is that after 240 minutes the drug would have worn off.
I believe bobpursley gave you that same answer in an earlier post, using Calculus.
    
is a quadratic function of the form you were given
f(x) = ax^2 + bx + c
a = -.0002
b = .06
c = 0
x = -b/(2a) gives you the x value where the max value of your function exists, so
t = -.06/(2(-.0002)) = 150
now sub that back into the equation to find the actual concentration
C(150) = .06(150) - .0002(150)^2 = 4.5 units
notice that t = 150 falls within the domain given of 0 ≤ t ≤ 240 , the reason probably is that after 240 minutes the drug would have worn off.
I believe bobpursley gave you that same answer in an earlier post, using Calculus.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.