Question

\(\max_{0 \leq x \leq \pi} \left( 16 \sin\left(\frac{x}{2}\right) \cos^3\left(\frac{x}{2}\right) \right)\) is equal to:

• A. 2
• B. \(3\sqrt{3}\)
• C. \(4\sqrt{3}\)
• D. \(6\sqrt{3}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the global maximum of a trigonometric function on a closed interval, we map the expression using substitution to simplify the domain, calculate the first derivative to find critical points, and substitute them back to yield the maximum.
Step 2: Key Formula or Approach:
For \(y = f(t)\), the maximum occurs where \(f'(t) = 0\).
Apply substitution \(t = x/2\).
Step 3: Detailed Explanation:
Let \(t = x/2\). As \(x \in [0, \pi]\), \(t \in [0, \pi/2]\).
In this interval, \(\cos t \ge 0\), so we can drop the absolute value.
Let \(f(t) = 16 \sin t \cos^3 t\).
Find the derivative using product and chain rules:
\[ f'(t) = 16 (\cos t \cdot \cos^3 t + \sin t \cdot 3\cos^2 t(-\sin t)) \] \[ f'(t) = 16(\cos^4 t - 3\sin^2 t \cos^2 t) = 16\cos^2 t(\cos^2 t - 3\sin^2 t) \] Set \(f'(t) = 0\):
Since \(t \in [0, \pi/2]\), \(\cos t \neq 0\) except at boundary \(\pi/2\) (which gives minimum 0).
\[ \cos^2 t - 3\sin^2 t = 0 \implies \tan^2 t = \frac{1}{3} \implies \tan t = \frac{1}{\sqrt{3}} \implies t = \frac{\pi}{6} \] Substitute \(t = \pi/6\) into \(f(t)\):
\[ f\left(\frac{\pi}{6}\right) = 16 \sin\left(\frac{\pi}{6}\right) \cos^3\left(\frac{\pi}{6}\right) = 16 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)^3 \] \[ = 8 \left(\frac{3\sqrt{3}}{8}\right) = 3\sqrt{3} \] Step 4: Final Answer:
The maximum value is \(3\sqrt{3}\).