Problem 64 - Algebra and Trigonometry

Question

Pharmaceuticals \(\quad\) 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 \(\mathrm{mg} / \mathrm{L}\) . 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: Identify the function type

We are given a quadratic function of the form \( C(t) = 0.06t - 0.0002t^2 \). This is a parabola that opens downwards because the coefficient of \( t^2 \) is negative.

Step 2: Find the vertex of the parabola

The maximum value of a downward-opening parabola occurs at its vertex. For a quadratic function given by \( ax^2 + bx + c \), the vertex occurs at \( t = -\frac{b}{2a} \). Here, \( a = -0.0002 \) and \( b = 0.06 \). Substitute these values in to find \( t \):\[t = -\frac{0.06}{2(-0.0002)} = 150.\]

Step 3: Calculate the maximum concentration

Substitute \( t = 150 \) back into the original equation \( C(t) = 0.06t - 0.0002t^2 \) to find the maximum concentration:\[C(150) = 0.06(150) - 0.0002(150)^2 = 9 - 4.5 = 4.5 \, \text{mg/L}.\]

Step 4: Verify the constraints

Check if \( t = 150 \) satisfies the problem's constraints \( 0 \leq t \leq 240 \). Since 150 is within this range, the calculation is valid.

Key Concepts

Understanding Parabolas

Imagine throwing a ball into the air. The path it follows resembles a curve called a parabola. Parabolas are the graphical representations of quadratic functions, typically taking the form of \( ax^2 + bx + c \). Depending on the sign of the leading coefficient (\( a \)), these curves can open upwards or downwards.

• If \( a > 0 \), the parabola opens upwards, looking like a U shape.
• If \( a < 0 \), it opens downwards, like an upside-down U.

For the given function \( C(t) = 0.06t - 0.0002t^2 \), since the coefficient of \( t^2 \) is negative \(-0.0002\), it forms a downward opening parabola. This shape indicates that the function has a peak point—also known as a maximum value.

The Maximum Value of a Quadratic Function

In all downward opening parabolas, the 'highest' point on the graph is the maximum value. This value is particularly significant in various applications, such as calculating the peak concentration of a drug in the bloodstream. This point, or vertex, is found exactly in the middle of the parabola, on the line of symmetry.

For the quadratic function \( ax^2 + bx + c \), the formula to find this important point, or maximum value, is determined by the vertex formula \( t = -\frac{b}{2a} \). Once you find \( t \), substituting it back into the original function will give you the maximum value. In this case:

• The time \( t = 150 \) when the concentration peaks.
• The peak concentration value is \( 4.5 \, \text{mg/L} \).

Applying the Vertex Formula

Finding the vertex of a parabola is crucial when working with quadratic functions. The vertex formula \( t = -\frac{b}{2a} \) provides the x-coordinate (or in this problem, the t-coordinate) of the vertex. Let's apply this to the current drug concentration scenario.

Substitute the known values of \( a = -0.0002 \) and \( b = 0.06 \) into the vertex formula:
\[t = -\frac{0.06}{2(-0.0002)} = 150.\]
This calculation shows exactly when the maximum concentration of the drug occurs in the bloodstream. Once the vertex is located, use this \( t \) value in the original function to determine the concentration value at this peak point:
\[C(150) = 0.06(150) - 0.0002(150)^2 = 4.5 \, \text{mg/L}.\] This approach helps not only in solving this problem but is a valuable tool for any quadratic function analysis.