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Geometry Difficulty 2.6 Junior Find the answer

ABCDABCD is a square with side of unit length. Points EE and FF are taken respectively on sides ABAB and ADAD so that AE=AFAE = AF and the quadrilateral CDFECDFE has maximum area. In square units this maximum area is:

Pick one

Solution

Let AE=AF=xAE=AF=x
[CDFE]=[ABCD]−[AEF]−[EBC]=1−x22−1−x2[CDFE]=[ABCD]-[AEF]-[EBC]=1-\frac{x^2}{2}-\frac{1-x}{2}
Or
[CDFE]=54−(x−12)22≤58[CDFE]=\frac{\frac{5}{4}-(x-\frac{1}{2})^2}{2}\le \frac{5}{8}
As (x−12)2≥0(x-\frac{1}{2})^2\ge 0
So [CDFE]≤58[CDFE]\le \frac{5}{8}
Equality occurs when AE=AF=x=12AE=AF=x=\frac{1}{2}
So the maximum value is 58\frac{5}{8}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.