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 area of triangle ABC is maximized, we can follow these steps:

### Step 1: Determine the coordinates of the vertices
The vertices of triangle ABC 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 for the area of a triangle given by its vertices:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
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:
\[ A = \frac{1}{2} \left| -\sin^3 \theta \cdot \cos \theta \right| = \frac{1}{2} \sin^3 \theta \cos \theta \]

### 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.
\[ A = \frac{1}{2} \sin^3 \theta \cos \theta \]
Using the product rule, we differentiate:
\[ \frac{dA}{d\theta} = \frac{1}{2} \left( 3\sin^2 \theta \cos \theta \cdot \frac{d(\sin \theta)}{d\theta} + \sin^3 \theta \cdot \frac{d(\cos \theta)}{d\theta} \right) \]
This gives:
\[ \frac{dA}{d\theta} = \frac{1}{2} \left( 3\sin^2 \theta \cos^2 \theta - \sin^3 \theta \sin \theta \right) \]
Simplifying further:
\[ \frac{dA}{d\theta} = \frac{1}{2} \left( 3\sin^2 \theta \cos^2 \theta - \sin^4 \theta \right) \]
Setting this equal to zero:
\[ 3\sin^2 \theta \cos^2 \theta - \sin^4 \theta = 0 \]
Factoring out \( \sin^2 \theta \):
\[ \sin^2 \theta (3\cos^2 \theta - \sin^2 \theta) = 0 \]

### Step 4: Solve for \( \theta \)
From \( \sin^2 \theta = 0 \), we get \( \theta = 0 \), which is not in the interval \( (0, \frac{\pi}{2}) \).

From \( 3\cos^2 \theta - \sin^2 \theta = 0 \):
\[ 3\cos^2 \theta = \sin^2 \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 3(1 - \sin^2 \theta) = \sin^2 \theta \]
\[ 3 - 3\sin^2 \theta = \sin^2 \theta \]
\[ 4\sin^2 \theta = 3 \implies \sin^2 \theta = \frac{3}{4} \implies \sin \theta = \frac{\sqrt{3}}{2} \]
Thus, \( \theta = \frac{\pi}{3} \).

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