Maximum Value of Trigonometric Function Product

To find the maximum value of the trigonometric expression $f(x) = 16 \sin\left(\frac{x}{2}\right) \cos^3\left(\frac{x}{2}\right)$ for $0 \le x \le \pi$, let us define a new variable $t = \frac{x}{2}$.

Since $0 \le x \le \pi$, the domain for $t$ becomes $0 \le t \le \frac{\pi}{2}$. The function can now be written as: $g(t) = 16 \sin(t) \cos^3(t)$

Method 1: Using Calculus

To find the critical points of $g(t)$ in the interval $\left[0, \frac{\pi}{2}\right]$, we differentiate $g(t)$ with respect to $t$: $g'(t) = 16 \left( \frac{d}{dt}[\sin(t)] \cos^3(t) + \sin(t) \frac{d}{dt}[\cos^3(t)] \right)$ $g'(t) = 16 \left( \cos(t) \cos^3(t) + \sin(t) \cdot 3\cos^2(t)(-\sin(t)) \right)$ $g'(t) = 16 \cos^2(t) \left( \cos^2(t) - 3\sin^2(t) \right)$

Setting $g'(t) = 0$:

1. $\cos^2(t) = 0 \implies t = \frac{\pi}{2}$
2. $\cos^2(t) - 3\sin^2(t) = 0 \implies \tan^2(t) = \frac{1}{3} \implies \tan(t) = \frac{1}{\sqrt{3}}$ (since $t \in \left[0, \frac{\pi}{2}\right]$)

From $\tan(t) = \frac{1}{\sqrt{3}}$, we get $t = \frac{\pi}{6}$.

Now, evaluating $g(t)$ at the boundary points and the critical point:

• At $t = 0$: $g(0) = 16 \sin(0) \cos^3(0) = 0$
• At $t = \frac{\pi}{2}$: $g\left(\frac{\pi}{2}\right) = 16 \sin\left(\frac{\pi}{2}\right) \cos^3\left(\frac{\pi}{2}\right) = 0$
• At $t = \frac{\pi}{6}$: $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ $g\left(\frac{\pi}{6}\right) = 16 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)^3 = 8 \times \frac{3\sqrt{3}}{8} = 3\sqrt{3}$

Method 2: Using the AM-GM Inequality

Consider the square of $g(t)$: $[g(t)]^2 = 256 \sin^2(t) \cos^6(t) = 256 \sin^2(t) \left(1 - \sin^2(t)\right)^3$

Let $y = \sin^2(t)$ where $0 \le y \le 1$. We maximize the expression $y(1-y)^3$. Multiplying by $3$: $3y(1-y)^3 = (3y)(1-y)(1-y)(1-y)$

Applying the AM-GM inequality to the non-negative terms $3y, (1-y), (1-y), (1-y)$: $\sqrt[4]{(3y)(1-y)^3} \le \frac{3y + (1-y) + (1-y) + (1-y)}{4} = \frac{3}{4}$

Raising both sides to the $4^{\text{th}}$ power: $3y(1-y)^3 \le \left(\frac{3}{4}\right)^4 = \frac{81}{256}$ $y(1-y)^3 \le \frac{27}{256}$

Substituting this back into $[g(t)]^2$: $[g(t)]^2 \le 256 \times \frac{27}{256} = 27$ $g(t) \le \sqrt{27} = 3\sqrt{3}$

Equality holds when $3y = 1 - y \implies y = \frac{1}{4} \implies \sin(t) = \frac{1}{2} \implies t = \frac{\pi}{6}$.

Thus, the maximum value is $3\sqrt{3}$.

Correct Answer: Option B ($3\sqrt{3}$)