We apply Pythagoras Theorem.

The diagonal of this rectangle is \(r+r=2r\)

\(x^2+y^2=(2r)^2\)

\(x^2+y^2=4r^2\)

\(y^2=4r^2-x^2\)

\(y=\sqrt{4r^2-x^2}\)

Let it be A-area of rectangle.

\(A=xy\)

\(A=x\sqrt{4r^2-x^2}\)

Now, we differentiate A(x)

\(A'(x)=(x\sqrt{4r^2-x^2})'\)

\(=x'\sqrt{4r^2-x^2}+(\sqrt{4r^2-x^2})'x\)

\(=1\cdot\sqrt{4r^2-x^2}+(-\frac{x}{\sqrt{4r^2-x^2}})\)

\(=\sqrt{4r^2-x^2}-\frac{x^2}{\sqrt{4r^2-x^2}}\)

\(=\frac{4r^2-x^2-x^2}{\sqrt{4r^2-x^2}}\)

\(=\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}\)

Solve \(A'(x)=0\)

\(\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}=0\)

\(4r^2-2x^2=0\)

\(x=\sqrt{2}r,x=-\sqrt{2}r\)

x must be positive, so \(x=\sqrt{2}r\)

Largest area is: (subtitute \(x=\sqrt{2}r\) in equation of area)

\(A=\sqrt{2}r\sqrt{4r^2-(\sqrt{2}r)^2}\)

\(A=\sqrt{2}r\cdot\sqrt{4r^2-2r^2}\)

\(A=\sqrt{2}r\cdot\sqrt{2r^2}\)

\(A=\sqrt{2}r\cdot\sqrt{2}r\)

\(A=2r^2\)