The concentration C(t), in milligrams per cubic centimetre, of a certain medicine in a patient's bloodstream is given by C(t)= (0.1t)/(t+3)^2 where t is number of hours after the medicine is taken. Determine the maximum and minimum concentrations between the first and sixth hours after the medicine is taken.

How did the book get answer max t=3 and min t=1

Find the derivative using the quotient rule

C'=(.1(t+3)^2-(2t+6)(.1t))/(t+3)^4
find where .1(t+3)^2 - (2t+6)(.1t) = 0
solving for t
t = 3
Evaluate C(1)
C(3) and C(6)
C(3) gives the maximum value, C(1) gives the minimum

Well, it seems that the book was trying to be sneaky and solve this math problem using calculus. But don't worry, I'm here to provide you with a more entertaining explanation.

Think of the concentration of the medicine as a roller coaster ride. The concentration C(t) is like the height of the roller coaster at different times t. And just like a roller coaster, we want to find the highest point (maximum) and the lowest point (minimum) between the first and sixth hours.

Now, imagine yourself on this medication-themed roller coaster. As you reach the top of the roller coaster, that would be the maximum concentration. So, we need to find the time when you're at the highest point on the ride. In this case, it's when t = 3. So the maximum concentration occurs after 3 hours.

On the other hand, as you go down the steep slopes of the roller coaster, that would be the lowest point (minimum) on the concentration graph. So we need to find the time when you're at the lowest point on the medication ride. In this case, it's when t = 1. So the minimum concentration occurs after 1 hour.

Remember, though, this is just a playful explanation. The book probably used some fancy math techniques to find these values, but I prefer to bring a little laughter into the equation.

To find the maximum and minimum concentrations between the first and sixth hours after the medicine is taken, we need to find the values of t that correspond to these extremes.

First, let's find the maximum concentration. To do this, we need to determine where the derivative of C(t) changes from positive to negative. The derivative gives us information about the rate of change of C(t).

1. Take the derivative of C(t) with respect to t:
C'(t) = [0.1*(t+3)^2 - 0.1*2*(t+3)*(t)] / (t+3)^4
Simplify:
C'(t) = [0.1(t+3 - 2t)] / (t+3)^3
C'(t) = -[0.2t - 0.6] / (t+3)^3

2. Set C'(t) equal to zero and solve for t:
-[0.2t - 0.6] / (t+3)^3 = 0
0.2t - 0.6 = 0
0.2t = 0.6
t = 0.6 / 0.2
t = 3

So, the maximum concentration occurs at t = 3 hours.

Next, let's find the minimum concentration. To do this, we need to check the endpoints of the given interval, which is between the first and sixth hours (1 ≤ t ≤ 6).

We evaluate the concentration function at the endpoints:
C(1) = (0.1*1) / (1+3)^2 = 0.01 / 16 = 0.000625 milligrams per cubic centimeter
C(6) = (0.1*6) / (6+3)^2 = 0.06 / 81 = 0.000741 milligrams per cubic centimeter

Comparing the two values, we can see that C(1) < C(6). Therefore, the minimum concentration occurs at t = 1 hour.

In summary, the maximum concentration occurs at t = 3 hours, and the minimum concentration occurs at t = 1 hour.

To determine the maximum and minimum concentrations, we need to find the critical points where the concentration function changes its behavior.

1. Maximum Concentration:
We find the critical points by taking the derivative of the concentration function, C(t), with respect to t. The critical points occur where the derivative is equal to zero or undefined.

C(t) = (0.1t) / (t + 3)^2

To find the derivative, we can use the quotient rule:

C'(t) = [ (t + 3)^2 * 0.1 - (0.1t) * 2(t + 3) ] / (t + 3)^4

Simplifying this expression, we get:

C'(t) = [ 0.1(t^2 + 6t + 9) - 0.2t(t + 3) ] / (t + 3)^4
= [ 0.1t^2 + 0.6t + 0.9 - 0.2t^2 - 0.6t ] / (t + 3)^4
= [ -0.1t^2 + 0.9 ] / (t + 3)^4

To find the critical points, we set the derivative equal to zero:

-0.1t^2 + 0.9 = 0

Solving this equation, we get:

-0.1t^2 = -0.9
t^2 = 9
t = ±3

Since we are interested in the times between the first and sixth hours, we only consider the positive root, t = 3. So, the maximum concentration occurs at t = 3.

2. Minimum Concentration:
To find the minimum concentration, we need to check the endpoints of the given time interval, which are the first and sixth hours.

For t = 1:

C(1) = (0.1 * 1) / (1 + 3)^2
= 0.1 / 16
= 0.00625 mg/cm^3

For t = 6:

C(6) = (0.1 * 6) / (6 + 3)^2
= 0.6 / 81
= 0.00741 mg/cm^3

Comparing the concentrations at t = 1 and t = 6, we see that the minimum concentration occurs at t = 1.

Therefore, according to the given concentration function, the maximum concentration occurs at t = 3 hours, and the minimum concentration occurs at t = 1 hour.