Answer
The rectangle of largest area that can be inscribed in a circle of radius $r$ is a square with a side of length $\sqrt{2}~r$
Work Step by Step
Let $x$ be the width of the rectangle and let $y$ be the length of the rectangle.
Note that $x \gt 0$ and $y \gt 0$
A diagonal across the rectangle has the length $2r$. Then:
$x^2+y^2 = (2r)^2$
$y = \sqrt{4r^2-x^2}$
We can write an expression for the area $A$:
$A = xy = x\sqrt{4r^2-x^2}$
We can find the point where $A'(x) = 0$:
$A(x) = x\sqrt{4r^2-x^2}$
$A'(x) = \sqrt{4r^2-x^2}-\frac{x^2}{\sqrt{4r^2-x^2}}= 0$
$\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}= 0$
$4r^2-2x^2 = 0$
$2x^2 = 4r^2$
$x^2 = 2r^2$
$x = \sqrt{2}~r$
When $0 \lt x \lt \sqrt{2}~r$, then $A'(x) \gt 0$
When $x \gt \sqrt{2}~r$, then $A'(x) \lt 0$
Thus, $x=\sqrt{2}~r$ is the point where the area $A$ is a maximum.
We can find $y$:
$y = \sqrt{4r^2-x^2}$
$y = \sqrt{4r^2-(\sqrt{2}~r)^2}$
$y = \sqrt{4r^2-2r^2}$
$y = \sqrt{2}~r$
The rectangle of largest area that can be inscribed in a circle of radius $r$ is a square with a side of length $\sqrt{2}~r$
