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  4. The ABCD is a square with sides of unit length. Points E and F are taken on 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

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Q. The $ABCD$ is a square with sides of unit length. Points $E$ and $F$ are taken on 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

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Solution:

Area of $CDFE$
image
$A=1-\frac{1}{2} x^{2}-\frac{1}{2}(1-x)$
$=\frac{2-x^{2}-1+x}{2}=\frac{1+x-x^{2}}{2}$
$\therefore A _{\max }=\frac{1+\frac{1}{2}-\frac{1}{4}}{2}=\frac{5}{8}$ at $x =\frac{1}{2}$