Problem 45 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Data show that the number of nonfarm, full-time, self-employed women can be approximated by $$ N(t)=0.81 t-1.14 \sqrt{t}+1.53 \quad(0 \leq t \leq 6) $$ where \(N(t)\) is measured in millions and \(t\) is measured in 5 -yr intervals, with \(t=0\) corresponding to the beginning of 1963\. Determine the absolute extrema of the function \(N\) on the interval \([0,6]\). Interpret your results.

Step-by-Step Solution

Answer

The absolute minimum number of nonfarm, full-time, self-employed women occurs at the beginning of 1963 with approximately 1.53 million women. The absolute maximum number occurs after 30 years (at the end of the given interval) with approximately 3.37 million women.

Step 1: Find the first derivative of N(t)

We are given the function \(N(t) = 0.81t - 1.14\sqrt{t} + 1.53\). To find the critical points, we need to compute the first derivative and then find when it is 0 or undefined: \[ N'(t) = \frac{d}{dt} (0.81t - 1.14\sqrt{t} + 1.53) \]

Step 2: Compute the first derivative using the power rule

Let's rewrite the function in terms of powers of t so we can use the power rule to compute the first derivative: \[ N(t) = 0.81t - 1.14t^{\frac{1}{2}} + 1.53 \] Now we can compute the first derivative using the power rule, which states that if \(f(t) = t^n\), then \(f'(t) = nt^{(n-1)}\): \[ N'(t) = 0.81\cdot(1)t^{(1-1)} - 1.14\cdot\frac{1}{2}t^{(\frac{1}{2}-1)} + 0 \] \[ N'(t) = 0.81 - 0.57t^{-\frac{1}{2}} \]

Step 3: Find when the first derivative is 0 or undefined

To find the critical points, we need to solve the equation \(N'(t) = 0\) and check for when the derivative is undefined: Solve for N'(t) = 0: \[ 0.81 - 0.57t^{-\frac{1}{2}} = 0 \] \[ 0.57t^{-\frac{1}{2}} = 0.81 \] \[ t^{-\frac{1}{2}} = \frac{0.81}{0.57} \] \[ t^{\frac{1}{2}} = \frac{0.57}{0.81} \] \[ t = \left(\frac{0.57}{0.81}\right)^2 = 0.495 \]

Step 4: Check the endpoints and critical points

Now we need to evaluate the function N(t) at the critical point and endpoints to compare them and find the absolute extrema: \[ N(0) = 0.81(0) - 1.14\sqrt{0} + 1.53 = 1.53 \] \[ N(6) = 0.81(6) - 1.14\sqrt{6} + 1.53 = 3.37 \] \[ N(0.495) = 0.81(0.495) - 1.14\sqrt{0.495} + 1.53 = 1.56 \]

Step 5: Conclusion and Interpretation

We can conclude that the absolute minimum occurs at \(t=0\), with a value of \(1.53\) million nonfarm, full-time, self-employed women. The absolute maximum occurs at \(t=6\), with a value of \(3.37\) million. Interpretation: The number of nonfarm, full-time, self-employed women was at its lowest at the beginning of 1963 with approximately 1.53 million women, and increased to approximately 3.37 million after 30 years (at the end of the given interval). There is a local maximum at approximately \(t = 0.495\), but it is lower than the maximum value at the end of the interval.

Key Concepts

Derivative Calculations

Derivative calculations play a crucial role in understanding how functions behave over certain intervals. In this problem, we focus on determining the derivative of the function \(N(t) = 0.81t - 1.14\sqrt{t} + 1.53\). The goal is to identify changes in the rate at which the number of nonfarm, full-time, self-employed women changes over time.
To calculate the derivative, we start by rewriting the function using powers of \(t\): \(N(t) = 0.81t - 1.14t^{\frac{1}{2}} + 1.53\).
This transformation allows us to efficiently use the power rule, which states that \(f(t) = t^n\) has a derivative of \(f'(t) = nt^{(n-1)}\).

• Differentiate \(0.81t\) to get \(0.81\), because the derivative of a constant times \(t^1\) is simply the constant.
• For \(-1.14t^{\frac{1}{2}}\), apply the power rule to get \(-1.14 \cdot \frac{1}{2} t^{-\frac{1}{2}}\), simplifying to \(-0.57t^{-\frac{1}{2}}\).
• Constant terms like \(+1.53\), disappear, because the derivative of any constant is 0.

After performing these calculations, the derivative is found to be \(N'(t) = 0.81 - 0.57t^{-\frac{1}{2}}\). This information is vital as it helps identify critical points where the function's growth rate changes significantly.

Critical Points Analysis

Critical points in calculus represent moments where a function's derivative is zero or undefined, often indicating potential extrema such as minima, maxima, or saddle points.
To achieve this, you begin by setting the derivative \(N'(t) = 0.81 - 0.57t^{-\frac{1}{2}}\) equal to zero and solving for \(t\).
This gives us the equation \(0.81 = 0.57t^{-\frac{1}{2}}\). Solving this:

• First, isolate the term with \(t\): \(t^{-\frac{1}{2}} = \frac{0.81}{0.57}\).
• Next, convert the negative exponent: \(t^{\frac{1}{2}} = \frac{0.57}{0.81}\).
• Finally, square both sides to solve for \(t\): \(t = \left(\frac{0.57}{0.81}\right)^2 = 0.495\).

After identifying \(t = 0.495\) as a critical point, the next step is to determine its nature by evaluating the function \(N(t)\) at this point and comparing it with values at the endpoints of the interval \([0,6]\).
Critical point analysis helps find opportunities where the function’s trend experiences significant changes, leading to further examination of absolute extrema.

Absolute Extrema Determination

Determining absolute extrema involves finding the highest and lowest values of a function on a closed interval.
In this exercise, we are asked to find the absolute extrema of \(N(t)\) on the interval \([0,6]\).
This process includes checking the function value at each endpoint and any critical points discovered.

• Evaluate \(N(t)\) at \(t = 0\): \(N(0) = 1.53\).
• Evaluate \(N(t)\) at \(t = 6\): \(N(6) = 3.37\).
• Evaluate \(N(t)\) at the critical point \(t = 0.495\): \(N(0.495) = 1.56\).

Comparing these values, we observe:

• The absolute minimum occurs at \(t = 0\) with \(N(0) = 1.53\), representing the lowest number of self-employed women at the start of 1963.
• The absolute maximum is located at \(t = 6\) with \(N(6) = 3.37\), showing the highest number after 30 years.

Through this analysis, we’ve determined that while there is a local maximum at \(t = 0.495\), it does not exceed the function’s value at the interval's endpoint \(t = 6\).
Such assessments are significant for understanding overall trends and changes over the given period.