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.
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?
1 answer