Problem 65 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The percentage of foreign-born residents in the United States from 1910 through 2000 is approximated by the function \(P(t)=0.04363 t^{3}-0.267 t^{2}-1.59 t+14.7 \quad(0 \leq t \leq 9)\) where \(t\) is measured in decades, with \(t=0\) corresponding to \(1910 .\) Show that the percentage of foreign-born residents was lowest in early 1970 . Hint: Use the quadratic formula.

Step-by-Step Solution

Answer

The percentage of foreign-born residents in the United States is given by the function \(P(t)=0.04363t^3-0.267t^2-1.59t+14.7\) for \(0 \leq t \leq 9\), where t is measured in decades with \(t=0\) corresponding to 1910. To find the minimum value, we calculate the derivative \(P'(t)=0.13089t^2-0.534t-1.59\) and set it to 0. Using the quadratic formula with \(a=0.13089\), \(b=-0.534\), and \(c=-1.59\), we find the critical point \(t \approx 5.2\), which corresponds to the early 1970s, specifically in early 1972. The second derivative test confirms that this is a minimum point, so the percentage of foreign-born residents was lowest in the early 1970s.

Step 1: Find the derivative of P(t)

First, we need to find the derivative of P(t) with respect to t. P(t) is a cubic function, and its derivative will be a quadratic function. Derivative of P(t) with respect to t: \[P'(t) = \frac{d}{dt} (0.04363 t^3 - 0.267 t^2 - 1.59 t +14.7)\] Applying the power rule to differentiate each term in the function: \[P'(t) = (3 \cdot 0.04363) t^2 - (2 \cdot 0.267) t - 1.59\] Simplifying the expression: \[P'(t) = 0.13089t^2 - 0.534t - 1.59\]

Step 2: Set the derivative equal to 0 to find critical points

To find minimum points of P(t), we need to set its derivative equal to 0 and solve for t. \[0.13089t^2 - 0.534t - 1.59 = 0\]

Step 3: Solve the quadratic equation using the quadratic formula

Now, we will use the quadratic formula to solve for the critical points of the quadratic equation we obtained in Step 2. The quadratic formula is: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] In our case, a = 0.13089, b = -0.534, and c = -1.59. Plugging these values into the formula, we get: \[t = \frac{0.534 \pm \sqrt{(-0.534)^2 - 4(0.13089)(-1.59)}}{2(0.13089)}\] Solving for the two potential t values. We find that one of them is outside our given range of 0 ≤ t ≤ 9, so we ignore this one. The valid t value is approximately 5.2. That means the rate at which the foreign-born population was decreasing was minimum in the early 1970's, precisely in early 1972 (1910 + 5.2*10). Note: To verify it's a minimum, we can take the second derivative of P(t), P''(t), and verify that it's positive for the critical point found (t ≈ 5.2). The second derivative is P''(t) = 0.26178t - 0.534, and for t ≈ 5.2, it is positive, confirming our result that the percentage of foreign-born residents was indeed lowest in the early 1970s.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in calculus and algebra for solving quadratic equations of the form \(ax^2 + bx + c = 0\). In calculus, it helps in finding the critical points of functions where the derivative is a quadratic equation. The quadratic formula is given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation. To utilize this formula effectively:

• Identify the coefficients \(a\), \(b\), and \(c\).
• Plug these values into the quadratic formula.
• Calculate the discriminant \(b^2 - 4ac\) and evaluate the square root.
• Solve for the potential solutions using the plus and minus sign \(\pm\).

The solutions will help find specific values related to various problems, such as determining critical points, as shown in the given exercise. Remember, only the solution(s) that fall within the specified range or constraints should be considered relevant to the problem.

Critical Points

Critical points in calculus are where a function's derivative is zero or undefined. These points are essential as they help identify potential maxima, minima, or points of inflection. To find critical points:

• Take the derivative of the function.
• Set the derivative equal to zero to form a new equation (this could be a quadratic equation, as in our exercise).
• Solve the equation for the variable to find critical points.

Understanding critical points is crucial in determining the behavior of a function. They indicate where significant changes in the increase or decrease of function behavior occur. In the context of our exercise: The derivative of the function \(P(t)\) was set to zero, resulting in a quadratic equation \(0.13089t^2 - 0.534t - 1.59 = 0\). Solving this using the quadratic formula gave us the values of \(t\), which are the critical points. The valid critical point found within the problem's constraints shows where the percentage of foreign-born residents reached its lowest in early 1970.

Second Derivative Test

The second derivative test in calculus helps determine if a critical point is a local minimum, maximum, or a point of inflection. Once a critical point \(c\) is found:

• Find the second derivative of the function.
• Evaluate the second derivative at the critical point \(c\).
• If \(f''(c) > 0\), the function has a local minimum at \(c\).
• If \(f''(c) < 0\), the function has a local maximum at \(c\).
• If \(f''(c) = 0\), the test is inconclusive, and the point might be a point of inflection.

For the given exercise, after determining the critical point \(t \approx 5.2\), the second derivative \(P''(t) = 0.26178t - 0.534\) was calculated. Evaluating this at \(t \approx 5.2\), we found that \(P''(t)\) is positive, verifying our critical point corresponds to a local minimum. Hence, this confirms the minimum percentage of foreign-born residents occurred in the early 1970s. Understanding these concepts can significantly aid in analyzing and interpreting the behavior of functions in various applications.