Let x and y be the length and breadth of a rectangle inscribed in a circle of radius r.
Then,
\(y^2+x^2=4r^2\)
= \(y = \sqrt{4r^2-x^2}\) ... (i)
If A be the area of rectangle then
A = x . y
For maxima or minima,
Hence, A is maximum when \(x =\sqrt{2r}\)
Putting \(x =\sqrt{2r}\) in (i) we get
\(y = \sqrt{4r^2-2r^2}\) \(=\sqrt{2r}\)
i,e., x = y = \(\sqrt{2r}\)
Therefore, area of rectangle is maximum when x = y
i.e., rectangle is a square.
