Let ABCD is a square with sides of unit length. Points E and F are taken om sides AB and AD respectively so that AE= AF. Let P be a point inside the square ABCD.The maximum possible area of quadrilateral CDFE is-

Options

A. `1//8`
B. `1//4`
C. `5//8`
D. `3//8`

Explanation

To solve the problem, we need to find the maximum possible area of quadrilateral CDFE inside square ABCD, where AE = AF. Let's denote AE = AF = x.

### Step-by-Step Solution:

1. **Define the Square and Points**:
- Let the vertices of square ABCD be:
- A(0, 1)
- B(1, 1)
- C(1, 0)
- D(0, 0)
- Points E and F are on sides AB and AD respectively:
- E(0, 1 - x)
- F(x, 1)

2. **Area of Quadrilateral CDFE**:
- The area of quadrilateral CDFE can be calculated using the formula:
\[ \text{Area of CDFE} = \text{Area of ABCD} - \text{Area of } \triangle AFE - \text{Area of } \triangle BCE \]

3. **Calculate Area of Square ABCD**:
- The area of square ABCD is:
\[ \text{Area of ABCD} = 1 \times 1 = 1 \text{ square unit} \]

4. **Calculate Area of Triangle AFE**:
- Triangle AFE is a right triangle with base AF and height AE:
\[ \text{Area of } \triangle AFE = \frac{1}{2} \times AE \times AF = \frac{1}{2} \times x \times x = \frac{x^2}{2} \]

5. **Calculate Area of Triangle BCE**:
- The base BE is (1 - x) and the height is 1:
\[ \text{Area of } \triangle BCE = \frac{1}{2} \times BE \times BC = \frac{1}{2} \times (1 - x) \times 1 = \frac{1 - x}{2} \]

6. **Combine Areas**:
- Now, substituting the areas into the equation for CDFE:
\[ \text{Area of CDFE} = 1 - \left(\frac{x^2}{2} + \frac{1 - x}{2}\right) \]
- Simplifying this gives:
\[ \text{Area of CDFE} = 1 - \left(\frac{x^2 + 1 - x}{2}\right) = 1 - \frac{x^2 - x + 1}{2} \]
- This can be rewritten as:
\[ A = \frac{2 - x^2 + x - 1}{2} = \frac{-x^2 + x + 1}{2} \]

7. **Maximize the Area**:
- To maximize the area A, we differentiate with respect to x:
\[ A' = \frac{1}{2}(-2x + 1) \]
- Setting the derivative to zero for critical points:
\[ -2x + 1 = 0 \implies x = \frac{1}{2} \]

8. **Check for Maximum**:
- To confirm it's a maximum, we check the second derivative:
\[ A'' = -1 < 0 \]
- Since A'' is negative, x = 1/2 is indeed a maximum.

9. **Calculate Maximum Area**:
- Substitute x = 1/2 back into the area formula:
\[ A\left(\frac{1}{2}\right) = \frac{-\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 1}{2} = \frac{-\frac{1}{4} + \frac{1}{2} + 1}{2} = \frac{-\frac{1}{4} + \frac{2}{4} + \frac{4}{4}}{2} = \frac{\frac{5}{4}}{2} = \frac{5}{8} \]

### Final Answer:
The maximum possible area of quadrilateral CDFE is \(\frac{5}{8}\).