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?
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.
I apologize for any confusion earlier. Let's try to answer your question step by step.
In this problem, you are given a quadratic function C(t) = 0.06t - 0.0002t^2, which represents the concentration of a drug in the patient's bloodstream after t minutes.
To find the maximum serum concentration, we need to determine the value of t when the function reaches its maximum value and then find the corresponding concentration.
Let's begin by rewriting the equation in the standard form for a quadratic function: C(t) = at^2 + bt + c. So, in this case, a = -0.0002, b = 0.06, and c = 0.
Now, we want to find the value of t when the drug concentration is at its maximum. In quadratic functions, the maximum or minimum point occurs at the vertex of the parabola. For a quadratic function in the form f(x) = ax^2 + bx + c, the x-coordinate of the vertex can be found using the formula x = -b / (2a).
In this case, we need to apply the same concept to find the value of t. The formula for t is t = -b / (2a) since our equation is in terms of t, not x.
Substituting the values of a and b into the formula, we have t = -(0.06) / (2 * (-0.0002)).
Now, let's calculate this value:
t = -(0.06) / (2 * (-0.0002))
t = 0.06 / 0.0004
t = 150
Therefore, the maximum serum concentration is reached after 150 minutes.
To find the maximum concentration, we need to substitute the value of t into the equation C(t) = 0.06t - 0.0002t^2.
C(150) = 0.06 * 150 - 0.0002 * 150^2
C(150) = 9 - 0.0002 * 22500
C(150) = 9 - 4.5
C(150) = 4.5
So, the maximum concentration is 4.5 mg/L.
Regarding the condition 0 ≤ t ≤ 240, it simply represents the time range within which we are considering the drug concentration. In this case, the drug concentration is only being analyzed between 0 and 240 minutes.
I hope this clarifies the problem for you. If you have any further questions or need additional assistance, please let me know.