Problem 67 - Algebra for College Students

Question

Solve each problem. The amount of nitrogen dioxide \(A\) in parts per million (ppm) that was present in the air in the city of Homer on a certain day in June is modeled by the function$$A(t)=-2 t^{2}+32 t+12$$where \(t\) is the number of hours after 6: 00 A.M. Use this function to find the time at which the nitrogen dioxide level was at its maximum.

Step-by-Step Solution

Answer

The nitrogen dioxide level is at its maximum at 2:00 P.M.

Step 1: Identify the function’s form

The given function is a quadratic equation in the form of \[A(t) = -2t^2 + 32t + 12 \] where the coefficient of the quadratic term (\(-2\)) is negative, indicating that it opens downwards and hence has a maximum point.

Step 2: Determine the vertex form

For a quadratic equation \[A(t) = at^2 + bt + c\], the time \(t\) at which the maximum or minimum value occurs is given by \( t = -\frac{b}{2a} \). Here, \(a = -2\) and \(b = 32\).

Step 3: Substitute values into the vertex formula

Substitute \(a = -2 \) and \(b = 32\) into the formula: t = -\frac{32}{2(-2)} = -\frac{32}{-4} = 8 Therefore, the maximum nitrogen dioxide level occurs at \(t=8\) hours after 6:00 A.M.

Step 4: Calculate the corresponding time

Since \(t = 8\) hours after 6:00 A.M., the maximum concentration of nitrogen dioxide occurs at \(6:00 \text{ A.M. } + 8 \text{ hours } = 2:00 \text{ P.M.} \).

Key Concepts

Quadratic Equations

A quadratic equation is a second-degree polynomial equation in one variable. It has the general form:
\[ax^2 + bx + c = 0\]
where:

• \(a\) is the coefficient of the squared term and must be non-zero.
• \(b\) is the coefficient of the linear term.
• \(c\) is the constant term.

Quadratic equations usually represent parabolas when graphed. Depending on the sign of \(a\):

• If \(a > 0\), the parabola opens upwards, and the vertex represents the minimum point.
• If \(a < 0\), the parabola opens downwards, and the vertex represents the maximum point.

In the given problem, the equation \(A(t) = -2t^2 + 32t + 12\) is a classic example of a quadratic equation with \(a = -2\), \(b = 32\), and \(c = 12\). Because \(a < 0\), we expect the function to have a maximum point.

Vertex Formula

The vertex of a parabola described by a quadratic equation \(ax^2 + bx + c\) is a crucial point. It can be computed using the vertex formula. For both maxima and minima, the x-value (or t-value in our problem) of the vertex is:
\[ t = -\frac{b}{2a} \]
This formula gives the value of the variable where the quadratic function attains its peak or valley. In our specific example, we identify:

• \(a = -2\)
• \(b = 32\)

By substituting these into the vertex formula: \[ t = -\frac{32}{2(-2)} = \frac{32}{4} = 8 \]
Thus, the nitrogen dioxide reaches its maximum at 8 hours after 6:00 A.M., which converts to 2:00 P.M.

Maxima and Minima

Maxima and minima refer to the highest and lowest points on a parabola, respectively. For a quadratic function, these points are always located at the vertex. The nature of the quadratic coefficient \(a\) determines whether we have a maximum or minimum:

• If \(a > 0\), the parabola opens upwards, and the vertex is the minimum point.
• If \(a < 0\), the parabola opens downwards, and the vertex is the maximum point.

In the equation \(A(t) = -2t^2 + 32t + 12\):

• \(a = -2\), so the parabola opens downwards.
• Thus, the vertex represents the maximum nitrogen dioxide level.

The vertex calculation provides this maximum level at a specific time, which we found occurs at 2:00 P.M. This concept is particularly useful in real-world applications where identifying peak values, such as maximum pollution levels, is crucial.