We apply Pythagoras Theorem.

The diagonal of this rectangle is r+r=2r

\(\displaystyle{x}^{{2}}+{y}^{{2}}={\left({2}{r}\right)}^{{2}}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}={4}{r}^{{2}}\)

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

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

Let it be A-area of rectangle.

\(\displaystyle{A}={x}{y}\)

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

Now, we differentiate A(x)

\(\displaystyle{A}'{\left({x}\right)}={\left({x}\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}'\right.}\)

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

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

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

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

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

Solve A'(x)=0

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

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

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

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

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

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

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

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

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

\(\displaystyle{A}={2}{r}^{{2}}\)

The diagonal of this rectangle is r+r=2r

\(\displaystyle{x}^{{2}}+{y}^{{2}}={\left({2}{r}\right)}^{{2}}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}={4}{r}^{{2}}\)

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

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

Let it be A-area of rectangle.

\(\displaystyle{A}={x}{y}\)

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

Now, we differentiate A(x)

\(\displaystyle{A}'{\left({x}\right)}={\left({x}\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}'\right.}\)

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

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

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

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

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

Solve A'(x)=0

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

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

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

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

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

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

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

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

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

\(\displaystyle{A}={2}{r}^{{2}}\)