Problem 40 - Thomas Calculus

Question

In Exercises \(37-40\) , find the function's absolute maximum and minimum values and say where they occur. $$ h(\theta)=3 \theta^{2 / 3}, \quad-27 \leq \theta \leq 8 $$

Step-by-Step Solution

Answer

The absolute maximum value is 27 at \(\theta = -27\), and the minimum value is 0 at \(\theta = 0\).

Step 1: Understand the Function and Interval

We need to find the absolute maximum and minimum values of the function \( h(\theta) = 3 \theta^{2/3} \) over the interval \( -27 \leq \theta \leq 8 \). This means we will evaluate the function at the endpoints of the interval and any critical points within it.

Step 2: Find the Derivative

To find critical points, we first need to calculate the derivative of \( h(\theta) \). The function is \( h(\theta) = 3 \theta^{2/3} \). Using the power rule, the derivative is \( h'(\theta) = 3 \cdot \frac{2}{3} \theta^{-1/3} = 2 \theta^{-1/3} = \frac{2}{\theta^{1/3}} \).

Step 3: Identify Critical Points

Critical points occur where the derivative is zero or undefined. The derivative \( h'(\theta) = \frac{2}{\theta^{1/3}} \) is undefined at \( \theta = 0 \). Thus, \( \theta = 0 \) is our critical point.

Step 4: Evaluate at Endpoints and Critical Points

We evaluate the function at \( \theta = -27 \), \( \theta = 0 \), and \( \theta = 8 \).- At \( \theta = -27 \): \( h(-27) = 3(-27)^{2/3} = 3\cdot 9 = 27 \).- At \( \theta = 0 \): \( h(0) = 3(0)^{2/3} = 0 \).- At \( \theta = 8 \): \( h(8) = 3(8)^{2/3} = 3\cdot 4 = 12 \).

Step 5: Compare Values to Find Extrema

Compare the calculated function values: 27, 0, and 12. - The largest value is 27 at \( \theta = -27 \).- The smallest value is 0 at \( \theta = 0 \).

Key Concepts

Critical Points

Understanding critical points is crucial in calculus optimization. A critical point occurs where the derivative of a function is either zero or undefined. These points often indicate where maximum or minimum values might occur, but they don't always guarantee it.

• Derivative zero: This happens when the slope of the tangent to the curve is flat, or horizontal. For example, in many cases, it can reflect a peak (maximum) or a valley (minimum).
• Derivative undefined: Sometimes, critical points take place because the derivative doesn't exist at that point. Take our function, and you'll see that the derivative is undefined at \( \theta = 0 \), meaning we need to test it as a potential critical point.

Finding critical points is the starting point to determine potential maximum or minimum values within a given interval.

Derivative Calculation

Calculating derivatives is a cornerstone in solving optimization problems in calculus. The derivative of a function provides the rate at which the function is changing at any given point.
For the function \( h(\theta) = 3\theta^{2/3} \), applying the power rule helps us find the derivative easily. The power rule is a shortcut that allows us to differentiate functions of the form \( x^n \) by bringing down the power in front and subtracting one from the exponent.

• Start with \( h(\theta)= 3 \theta^{2/3} \).
• The derivative \( h'(\theta) = 2 \theta^{-1/3} \) is derived by applying the power rule.

This derivative tells us that the function's rate of change depends on the value of \( \theta \). At certain values, like \( \theta = 0 \), the derivative is undefined, signaling a potential critical point we must examine.

Absolute Maximum and Minimum

Seeking the absolute maximum and minimum within an interval is like identifying the tallest peak or the deepest valley along a trail. It requires considering all potential highs and lows within the specified boundaries.
To achieve this:

• Determine all critical points within the interval. This involves using the derivative to spot where the slope flattens or the derivative is undefined.
• Evaluate the function at these critical points and at the endpoints of the interval \( -27 \leq \theta \leq 8 \).

In our example, we evaluated \( h(\theta) = 3 \theta^{2/3} \) at \( \theta = -27, 0, \) and \( 8 \). Calculating these values:

• \( \theta = -27 \) gives an output of \( 27 \), which is the largest, hence the absolute maximum.
• \( \theta = 0 \) yields \( 0 \), the smallest output, marking the absolute minimum.
• \( \theta = 8 \) results in \( 12 \), a value between the other two.

By comparing these results, we easily identify where the absolute extremes occur, providing a comprehensive visualization of the function's behavior over the interval.