Problem 64 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the relative maxima and relative minima, if any, of each function. $$ F(t)=3 t^{5}-20 t^{3}+20 $$

Step-by-Step Solution

Answer

The function \(F(t) = 3t^5 - 20t^3 + 20\) has a relative maximum at \(t = -2\) with a value of F(-2) = -24, and a relative minimum at \(t = 2\) with a value of F(2) = 24.

Step 1: Find the first derivative F'(t)

Differentiate F(t) with respect to t: \( \\ F(t) = 3t^5 - 20t^3 + 20\\ F'(t) = \frac{d}{dt} (3t^5 - 20t^3 + 20)\\ F'(t) = 15t^4 - 60t^2 \)

Step 2: Find the critical points

To find the critical points, set F'(t) equal to zero and solve for t: \(\\ F'(t) = 15t^4 - 60t^2 = 0\\ 15t^4 - 60t^2 = 0 \\ 5t^2(3t^2 - 12) = 0 \) The critical points for this function are found when we solve this equation. So, \(\\ t^2=0 \Rightarrow t=0\\ 3t^2 = 12 \Rightarrow t^2=4 \Rightarrow t = \pm 2 \) The critical points are t = -2, 0, and 2.

Step 3: Analyze critical points using the first derivative test

Now, we'll use the first derivative test to determine the type and location of the relative extremum. For this, we'll analyze the intervals created by the critical points. - Interval 1: \(-\infty < t < -2\) - Interval 2: \(-2 < t < 0\) - Interval 3: \(0 < t < 2\) - Interval 4: \(2 < t < \infty\) We'll now analyze the sign of F'(t) in these intervals: - Interval 1 (\(-\infty < t < -2\)): Choose a test value, say t=-3. F'(-3) = 15*(-3)^4 - 60*(-3)^2 = 315 > 0. So, F'(t) > 0 on this interval, which means F(t) is increasing. - Interval 2 (\(-2 < t < 0\)): Choose a test value, say t=-1. F'(-1) = 15*(-1)^4 - 60*(-1)^2 = -45 < 0. So, F'(t) < 0 on this interval, which means F(t) is decreasing. - Interval 3 (\(0 < t < 2\)): Choose a test value, say t=1. F'(1) = 15*(1)^4 - 60*(1)^2 = -45 < 0. So, F'(t) < 0 on this interval, which means F(t) is decreasing. - Interval 4 (\(1 < t < \infty\)): Choose a test value, say t=3. F'(3) = 15*(3)^4 - 60*(3)^2 = 315 > 0. So, F'(t) > 0 on this interval, which means F(t) is increasing.

Step 4: Determine the relative maxima and minima

Combine the information from Step 3: - For t = -2, F'(t) changes from positive to negative. So, F(-2) = 3*(-2)^5 - 20*(-2)^3 + 20 = -24. This is a relative maximum. - For t = 0, F'(t) changes from negative to negative. So, there is no relative extrema at this point. - For t = 2, F'(t) changes from negative to positive. So, F(2) = 3*(2)^5 - 20*(2)^3 + 20 = 24. This is a relative minimum. Therefore, the relative maxima and minima of the function are: - Relative maximum at t = -2, with a value of F(-2) = -24. - Relative minimum at t = 2, with a value of F(2) = 24.

Key Concepts

First Derivative Test

The First Derivative Test is a fundamental tool in calculus used to determine the relative maxima and minima of a function. When you have a continuous function, the test helps assess where the function is increasing or decreasing by looking at the first derivative.
To apply the First Derivative Test, you first find the critical points of the function by finding where its derivative equals zero or is undefined. These points are crucial because they can represent local high or low points in the function's graph.
For each interval divided by these critical points, evaluate the sign of the first derivative:

• If the derivative changes from positive to negative, the function has a relative maximum.
• If it changes from negative to positive, the function has a relative minimum.
• If it does not change sign, there is no local extremum.

This test is particularly useful because it doesn't just tell you where the high and low points are, but also provides insight into the behavior of the function across different intervals.

Critical Points

Finding critical points is a step in calculus where you identify potential turning points in a function's graph. These points are found by setting the first derivative of the function equal to zero or where the derivative does not exist.
For the function \(F(t) = 3t^5 - 20t^3 + 20\), the first derivative is \(F'(t) = 15t^4 - 60t^2\).
To find its critical points, set the derivative equal to zero:\[15t^4 - 60t^2 = 0\]This equation can be factored to get \(5t^2(3t^2 - 12) = 0\). Solving this gives critical points at \(t = -2, 0,\) and \(2\).
Critical points are essential for analyzing the function's structure because they indicate where the function's growth might change direction, leading to possible local maxima, minima, or points where no relative extrema exist.

Relative Maximum and Minimum

Relative maxima and minima are the peaks and troughs in the graph of a function at certain points, not necessarily the highest or lowest points overall. They are part of a function's local behavior.
After identifying critical points, use the First Derivative Test to classify them:

• A point \(t = a\) is a relative maximum if the function goes from increasing to decreasing at that point. For \(F(t) = 3t^5 - 20t^3 + 20\), \(t = -2\) is a relative maximum with a value of \(-24\).
• A point \(t = b\) is a relative minimum if the function goes from decreasing to increasing at that point. The point \(t = 2\) is a relative minimum with a value of \(24\).

These points are significant, particularly in optimization problems, as they illustrate the local highs and lows of a function, guiding you in maximizing or minimizing a situation represented by the function.