# Find the dimensions of the rectangle of largest area that

Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius .

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