Find the maximum area of a rectangle that can be inscribed in a circle of radius $r$.

$2r^2$

The correct answer is Option (2) → $2r^2$ ##

The diagonal of the rectangle is the diameter of the circle, which is $2r$. Let the sides of the rectangle be $x$ and $y$. From the Pythagorean theorem:

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

The area of the rectangle is $A = x \times y$.

Express $A$ in terms of one variable, solve for $y$ from the equation $x^2 + y^2 = 4r^2$:

$ y^2 = 4r^2 - x^2 $

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

Therefore, the area becomes:

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

Differentiate $A(x)$ to find the critical points:

$ \frac{dA}{dx} = \sqrt{4r^2 - x^2} - \frac{x^2}{\sqrt{4r^2 - x^2}} $

Set $\frac{dA}{dx} = 0$ and solve:

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

After simplifying, $x = \sqrt{2}r$

At $x = \sqrt{2}r$ substitute into the area formula:

$ A = \sqrt{2}r \times \sqrt{4r^2 - 2r^2} = \sqrt{2}r \times \sqrt{2}r = 2r^2 $

The maximum area of the rectangle is $2r^2$.