Math Problem Statement

Find the maximum and minimum values of the function g(θ) = 6θ - 8sin(θ) on the interval [0, π].

Solution

We are tasked with finding the maximum and minimum values of the function g(θ)=6θ8sin(θ)g(\theta) = 6\theta - 8\sin(\theta) on the interval [0,π][0, \pi].

Steps to solve:

  1. Differentiate the function: To find the critical points, we need the derivative of g(θ)g(\theta). g(θ)=ddθ(6θ8sin(θ))=68cos(θ)g'(\theta) = \frac{d}{d\theta} \left( 6\theta - 8\sin(\theta) \right) = 6 - 8\cos(\theta)

  2. Set the derivative equal to zero to find critical points: 68cos(θ)=06 - 8\cos(\theta) = 0 cos(θ)=68=0.75\cos(\theta) = \frac{6}{8} = 0.75 Solving θ=cos1(0.75)\theta = \cos^{-1}(0.75), we get: θ=cos1(0.75)0.7227\theta = \cos^{-1}(0.75) \approx 0.7227 So, one critical point is θ0.7227\theta \approx 0.7227.

  3. Evaluate the function at the endpoints θ=0\theta = 0 and θ=π\theta = \pi, and at the critical point θ=0.7227\theta = 0.7227:

    • At θ=0\theta = 0: g(0)=6(0)8sin(0)=0g(0) = 6(0) - 8\sin(0) = 0
    • At θ=π\theta = \pi: g(π)=6π8sin(π)=6π18.8496g(\pi) = 6\pi - 8\sin(\pi) = 6\pi \approx 18.8496
    • At θ=0.7227\theta = 0.7227: g(0.7227)=6(0.7227)8sin(0.7227)g(0.7227) = 6(0.7227) - 8\sin(0.7227) Using a calculator: g(0.7227)4.3368(0.6614)4.3365.291=0.955g(0.7227) \approx 4.336 - 8(0.6614) \approx 4.336 - 5.291 = -0.955
  4. Determine the maximum and minimum values:

    • Minimum value: g(0.7227)0.955g(0.7227) \approx -0.955
    • Maximum value: g(π)18.8496g(\pi) \approx 18.8496

Thus, the minimum value is approximately -0.955, and the maximum value is approximately 18.8496.

Would you like more details or have any questions?

Related questions:

  1. What are critical points and why do we use them to find extreme values?
  2. How is the derivative used to identify increasing and decreasing behavior of a function?
  3. Why is it necessary to evaluate the function at the endpoints of the interval?
  4. Can we solve this problem graphically to verify our result?
  5. How would the solution change if the interval was different, for example, [0,2π][0, 2\pi]?

Tip:

When solving optimization problems on a closed interval, always check the function at the endpoints as well as at any critical points.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometry
Optimization

Formulas

g'(θ) = 6 - 8cos(θ)
Critical points are found by solving g'(θ) = 0
Evaluate g(θ) at critical points and interval endpoints

Theorems

Extreme Value Theorem
Derivative Test for Local Extrema

Suitable Grade Level

College-level Calculus