Statement 1: ABC is given triangle having respective sides a,b,c. D,E,F are points of the sides BC,CA,AB respectively so that AFDE is a parallelogram. The maximum area of the parallelogram is 1/4 bcsinA. Statement 2 : Maximum value of `2kx--x^2` is at x = k.

Options

A. a. `(1)/(4) " bcsinA "`
B. b. ` (1)/(2) " bcsin A "`
C. c. `" bcsin A "`
D. d. `(1)/(8) " bcsin A "`

Answer

To solve the problem, we will analyze both statements step by step.

### Statement 1:
**Given:** Triangle ABC with sides a, b, c. Points D, E, F are on sides BC, CA, AB respectively, forming parallelogram AFDE. We need to find the maximum area of the parallelogram.

1. **Understanding the Area of the Parallelogram:**
The area \( A \) of parallelogram AFDE can be expressed in terms of the base and height. The area can be calculated as:
\[ A = \text{base} \times \text{height} \]
Here, we can consider the base as \( AF \) and the height as the perpendicular distance from point E to line AF.

2. **Using the Sine Rule:**
The area of triangle ABC can be expressed using the formula:
\[ \text{Area}_{\triangle ABC} = \frac{1}{2}bc \sin A \]
Since AFDE is a parallelogram, its area will be related to the area of triangle ABC.

3. **Expressing the Area of the Parallelogram:**
The area of parallelogram AFDE can be expressed as:
\[ A_{AFDE} = 2 \times \text{Area}_{\triangle AEF} \]
Using the triangle area formula, we have:
\[ A_{AFDE} = 2 \times \left( \frac{1}{2} \times AF \times h \right) = AF \times h \]

4. **Maximizing the Area:**
To maximize the area, we can express \( AF \) and \( h \) in terms of a variable \( x \) that represents the ratio of the segments on the sides. After some algebra, we find:
\[ A_{AFDE} = cb \sin A \cdot \left( x(1 - x) \right) \]
where \( x \) is a variable that represents the proportion of the triangle's height.

5. **Finding the Maximum Area:**
To find the maximum area, we differentiate the area function with respect to \( x \) and set the derivative to zero:
\[ \frac{dA}{dx} = cb \sin A \cdot (1 - 2x) = 0 \]
Solving for \( x \) gives:
\[ x = \frac{1}{2} \]

6. **Calculating the Maximum Area:**
Substituting \( x = \frac{1}{2} \) back into the area equation:
\[ A_{max} = cb \sin A \cdot \left( \frac{1}{2} \left( 1 - \frac{1}{2} \right) \right) = \frac{1}{4} bc \sin A \]

Thus, the maximum area of the parallelogram AFDE is:
\[ \frac{1}{4} bc \sin A \]

### Statement 2:
**Given:** The maximum value of \( 2kx - x^2 \) occurs at \( x = k \).

1. **Understanding the Function:**
The function can be rewritten as:
\[ f(x) = -x^2 + 2kx \]

2. **Finding the Maximum:**
To find the maximum value, we differentiate \( f(x) \):
\[ f'(x) = -2x + 2k \]
Setting the derivative to zero gives:
\[ -2x + 2k = 0 \implies x = k \]

3. **Conclusion:**
The maximum value of the function \( 2kx - x^2 \) occurs at \( x = k \).

### Final Conclusion:
Both statements are correct:
- Statement 1: The maximum area of the parallelogram is \( \frac{1}{4} bc \sin A \).
- Statement 2: The maximum value of \( 2kx - x^2 \) occurs at \( x = k \).