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?
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.
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)
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?
To find the maximum serum concentration and the time at which it is reached, we need to determine the vertex of the quadratic function C(t) = 0.06t - 0.0002t^2.
The vertex of a quadratic function in the form f(t) = at^2 + bt + c can be found using the formula:
t = -b / (2a)
In our case, a = -0.0002 and b = 0.06.
Substituting the values into the formula:
t = -0.06 / (2 * -0.0002)
t = -0.06 / (-0.0004)
t = 150
So, the maximum serum concentration is reached at 150 minutes.
To find the maximum concentration, substitute the value of t = 150 into the quadratic function C(t):
C(150) = 0.06(150) - 0.0002(150)^2
C(150) = 9 - 0.0002(22500)
C(150) = 9 - 4.5
C(150) = 4.5
Therefore, the maximum serum concentration is 4.5 mg/L.