The maximum value of the expression `sin theta cos^(2)theta(AA theta in [0, pi])` is

Options

A. `(2)/(3)`
B. `(2)/(sqrt3)`
C. `(2)/(3sqrt3)`
D. `(1)/(sqrt3)`

Answer

To find the maximum value of the expression \( f(\theta) = \sin \theta \cos^2 \theta \) for \( \theta \in [0, \pi] \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(\theta) \):
\[ f'(\theta) = \frac{d}{d\theta} (\sin \theta \cos^2 \theta) \]
Using the product rule \( (uv)' = u'v + uv' \), where \( u = \sin \theta \) and \( v = \cos^2 \theta \):
- \( u' = \cos \theta \)
- \( v' = 2\cos \theta (-\sin \theta) = -2\sin \theta \cos \theta \)

Thus,
\[ f'(\theta) = \cos \theta \cdot \cos^2 \theta + \sin \theta \cdot (-2\sin \theta \cos \theta) \]
This simplifies to:
\[ f'(\theta) = \cos^3 \theta - 2\sin^2 \theta \cos \theta \]

### Step 2: Set the derivative to zero
To find the critical points, set \( f'(\theta) = 0 \):
\[ \cos^3 \theta - 2\sin^2 \theta \cos \theta = 0 \]
Factoring out \( \cos \theta \):
\[ \cos \theta (\cos^2 \theta - 2\sin^2 \theta) = 0 \]
This gives us two cases:
1. \( \cos \theta = 0 \)
2. \( \cos^2 \theta - 2\sin^2 \theta = 0 \)

### Step 3: Solve for critical points
**Case 1:** \( \cos \theta = 0 \)
This occurs at \( \theta = \frac{\pi}{2} \).

**Case 2:** \( \cos^2 \theta - 2\sin^2 \theta = 0 \)
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ 1 - \sin^2 \theta - 2\sin^2 \theta = 0 \implies 1 - 3\sin^2 \theta = 0 \implies \sin^2 \theta = \frac{1}{3} \]
Thus, \( \sin \theta = \pm \frac{1}{\sqrt{3}} \). Since \( \theta \) is in \( [0, \pi] \), we take \( \sin \theta = \frac{1}{\sqrt{3}} \).

### Step 4: Calculate corresponding \( \theta \)
Using \( \sin \theta = \frac{1}{\sqrt{3}} \):
\[ \theta = \arcsin\left(\frac{1}{\sqrt{3}}\right) \]

### Step 5: Evaluate \( f(\theta) \) at critical points
1. For \( \theta = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) \cos^2\left(\frac{\pi}{2}\right) = 1 \cdot 0 = 0 \]

2. For \( \sin \theta = \frac{1}{\sqrt{3}} \):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus,
\[ f\left(\theta\right) = \frac{1}{\sqrt{3}} \cdot \frac{2}{3} = \frac{2}{3\sqrt{3}} \]

### Step 6: Conclusion
The maximum value of \( f(\theta) \) occurs at \( \theta \) where \( \sin \theta = \frac{1}{\sqrt{3}} \):
\[ \text{Maximum value} = \frac{2}{3\sqrt{3}} \]

### Final Answer
The maximum value of the expression \( \sin \theta \cos^2 \theta \) for \( \theta \in [0, \pi] \) is \( \frac{2}{3\sqrt{3}} \).

---