If `A(0, 0), B(theta, cos theta) and C(sin^(3) theta, 0)` are the vertices of a triangle Abc, then the value of `theta` for which the triangle has the maximum area is `("where "theta in (0, (pi)/(2)))`

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Answer

To find the value of \( \theta \) for which the triangle \( ABC \) has the maximum area, we can follow these steps:

### Step 1: Determine the vertices of the triangle
The vertices of the triangle are given as:
- \( A(0, 0) \)
- \( B(\theta, \cos \theta) \)
- \( C(\sin^3 \theta, 0) \)

### Step 2: Use the formula for the area of a triangle
The area \( A \) of triangle \( ABC \) can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points \( A \), \( B \), and \( C \):
\[ A = \frac{1}{2} \left| 0(\cos \theta - 0) + \theta(0 - 0) + \sin^3 \theta(0 - \cos \theta) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| -\sin^3 \theta \cdot \cos \theta \right| = \frac{1}{2} \sin^3 \theta \cdot \cos \theta \]

### Step 3: Maximize the area function
To find the maximum area, we need to differentiate \( A \) with respect to \( \theta \) and set the derivative to zero:
\[ A = \frac{1}{2} \sin^3 \theta \cdot \cos \theta \]
Let \( A = \frac{1}{2} f(\theta) \) where \( f(\theta) = \sin^3 \theta \cdot \cos \theta \).

### Step 4: Differentiate \( f(\theta) \)
Using the product rule:
\[ f'(\theta) = 3\sin^2 \theta \cos \theta \cdot \cos \theta + \sin^3 \theta \cdot (-\sin \theta) \]
This simplifies to:
\[ f'(\theta) = 3\sin^2 \theta \cos^2 \theta - \sin^4 \theta \]
Setting \( f'(\theta) = 0 \):
\[ 3\sin^2 \theta \cos^2 \theta - \sin^4 \theta = 0 \]
Factoring out \( \sin^2 \theta \):
\[ \sin^2 \theta (3\cos^2 \theta - \sin^2 \theta) = 0 \]

### Step 5: Solve for \( \theta \)
From \( \sin^2 \theta = 0 \), we get \( \theta = 0 \).
From \( 3\cos^2 \theta - \sin^2 \theta = 0 \):
\[ 3\cos^2 \theta = \sin^2 \theta \implies \tan^2 \theta = 3 \implies \tan \theta = \sqrt{3} \]
Thus, \( \theta = \frac{\pi}{3} \).

### Step 6: Determine the maximum area condition
Since \( \theta \) must be in the interval \( (0, \frac{\pi}{2}) \), we check the values:
- \( \theta = 0 \) gives area = 0.
- \( \theta = \frac{\pi}{3} \) is valid and gives a maximum area.

### Conclusion
The value of \( \theta \) for which the triangle has the maximum area is:
\[ \theta = \frac{\pi}{3} \]

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