Take the derivative,set equal to zero, solve for time.
C'=0=.06-.0004t
t= solve.
Then put that t in the C(t) to find max concentration.
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?
3 answers
Ooops. I just realized you have not had calculus yet.\
Here is your way.
solve for the roots (when C=0)
0=.06t-.0002t^2=t(.06-.0002t)
or t=0, and t= .06/.0002
because the max has to be at the midpoint of these two roots (qudratics plot symettrically), the the max occurs between 0 and .06/.0002 or .06/.004. Use that time to solve for C(t)
Here is your way.
solve for the roots (when C=0)
0=.06t-.0002t^2=t(.06-.0002t)
or t=0, and t= .06/.0002
because the max has to be at the midpoint of these two roots (qudratics plot symettrically), the the max occurs between 0 and .06/.0002 or .06/.004. Use that time to solve for C(t)
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?
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