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(D) Let the rectangle be inscribed in a circle of radius $r$. Let the sides of the rectangle be $2x$ and $2y$. Since the rectangle is inscribed in a c

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$2 r^{2}$

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(D) Let the rectangle be inscribed in a circle of radius $r$. Let the sides of the rectangle be $2x$ and $2y$. Since the rectangle is inscribed in a circle,the diagonal of the rectangle is the diameter of the circle,so $(2x)^2 + (2y)^2 = (2r)^2$,which simplifies to $x^2 + y^2 = r^2$,or $y = \sqrt{r^2 - x^2}$.The area $A$ of the rectangle is $A = (2x)(2y) = 4x\sqrt{r^2 - x^2}$.To find the maximum area,differentiate $A$ with respect to $x$:$\frac{dA}{dx} = 4 \left( \sqrt{r^2 - x^2} + x \cdot \frac{1}{2\sqrt{r^2 - x^2}} \cdot (-2x) \right) = 4 \left( \frac{r^2 - x^2 - x^2}{\sqrt{r^2 - x^2}} \right) = \frac{4(r^2 - 2x^2)}{\sqrt{r^2 - x^2}}$.Setting $\frac{dA}{dx} = 0$ gives $r^2 - 2x^2 = 0$,so $x = \frac{r}{\sqrt{2}}$.Substituting $x = \frac{r}{\sqrt{2}}$ into the area formula:$A = 4 \left( \frac{r}{\sqrt{2}} \right) \sqrt{r^2 - \frac{r^2}{2}} = 4 \left( \frac{r}{\sqrt{2}} \right) \left( \frac{r}{\sqrt{2}} \right) = 4 \cdot \frac{r^2}{2} = 2r^2$.Thus,the maximum area is $2r^2$.

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is:

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