Problem 68 - Algebra and Trigonometry

Question

Pharmaceuticals When a certain drug is taken orally, the concentration of the drug in the patient's bloodstream after \(t\) minutes is given by \(C(t)=0.06 t-0.0002 t^{2},\) where \(0 \leq t \leq 240\) and the concentration is measured in mg/L. When is the maximum serum concentration reached, and what is that maximum concentration?

Step-by-Step Solution

Answer

The maximum concentration of 4.5 mg/L is reached at 150 minutes.

Step 1: Understand the Problem

We have a function that describes the concentration of a drug in the bloodstream over time: \(C(t) = 0.06t - 0.0002t^2\). We need to find when the maximum concentration occurs and what that maximum concentration is within the time frame \(0 \leq t \leq 240\).

Step 2: Find the Critical Points

To find the time at which the concentration is maximized, we first need to find the critical points by taking the derivative of \(C(t)\) and setting it equal to zero. The derivative is \(C'(t) = 0.06 - 0.0004t\). Set \(C'(t) = 0\) to find the critical points: \ \(0.06 - 0.0004t = 0\) \ Solve for \(t\): \ \(0.0004t = 0.06\) \ \(t = \frac{0.06}{0.0004} = 150\).

Step 3: Confirm Maximum with Second Derivative Test

Take the second derivative of \(C(t)\), \(C''(t) = -0.0004\). Since \(C''(t) < 0\), the function is concave down at \(t = 150\), confirming a local maximum at \(t = 150\).

Step 4: Compute the Maximum Concentration

Substitute \(t = 150\) back into the original function to find the maximum concentration: \ \(C(150) = 0.06(150) - 0.0002(150)^2 = 9 - 4.5 = 4.5\). The maximum concentration of the drug is 4.5 mg/L.

Step 5: Check the Endpoint Values

Verify that the concentration at the endpoints of the interval \([0, 240]\) are not higher. Calculate \(C(0) = 0.06(0) - 0.0002(0)^2 = 0\) and \(C(240) = 0.06(240) - 0.0002(240)^2 = 14.4 - 11.52 = 2.88\). Both are lower than the maximum found at \(t = 150\).

Key Concepts

Understanding the Derivative Test

In calculus, the derivative test is an essential tool for optimization.The primary purpose of this test is to find the maxima and minima of functions.When we have a function that models some real-life scenario, like the concentration of a drug over time, using a derivative helps us understand the rates of change and points where the rate stops changing.
To apply the first derivative test, we look for critical points by setting the derivative of the function equal to zero.This is because at critical points, the function's slope is zero, which means the function could be at a local maximum, local minimum, or even a saddle point.

• Take the derivative of the function.
• Set the derivative equal to zero.
• Solve for the variable to find critical points.

In our problem, the derivative of the concentration function is given by\(C'(t) = 0.06 - 0.0004t\).Setting this to zero finds the critical point, showing us when the maximum concentration might happen. Understanding this concept helps us make decisions about maximizing or minimizing functions effectively.

Identifying Critical Points

Critical points are values in the domain of the function where the derivative is zero or undefined.These points are important because they are where peaks and valleys, or flat spots like plateaus, might occur in a function.
In optimization problems, like determining when a drug reaches its peak concentration, identifying critical points helps us focus on the moments that need further investigation.In our example, by solving \(0.06 - 0.0004t = 0\), we find that \(t = 150\) is a critical point.

• Calculating the derivative \(C'(t) = 0.06 - 0.0004t\).
• Setting the derivative equal to zero gives \(t = 150\).
• This critical point needs further analysis via the second derivative test.

Notably, not all critical points are maximum or minimum points of function.Therefore, further tests are necessary to classify the nature of these points.

Explaining Concavity and Its Role

Concavity describes how a graph curves.A function can be concave upwards (like a bowl) or concave downwards (like an arch).The second derivative of a function tells us about its concavity:
- If the second derivative is positive at a point, the function is concave up.- If the second derivative is negative, the function is concave down.- If the second derivative is zero, the test is inconclusive.In our example, the second derivative, \(C''(t) = -0.0004\), tells us that the function is concave down.Since the second derivative is stable and negative, just like an arch, it shows that the point at \(t = 150\) is indeed a local maximum.

• The second derivative \(C''(t) = -0.0004\) indicates concavity.
• Since it is negative, the graph is concave down at that point.
• This confirms that our critical point is a maximum.

Concavity helps us conclude if a critical point genuinely indicates a peak or trough in the function, further guiding optimization efforts.