To solve the classic problem of finding the maximum area of a rectangle inscribed in a semicircle, follow these steps:

Problem Setup

Assume a semicircle of radius (r) centered at the origin, with equation (x^2 + y^2 = r^2) ((y \geq 0)). The rectangle is symmetric about the (y)-axis:

• Let the right corner of the rectangle on the semicircle be ((x, y)).
• Width of the rectangle: (2x) (since it spans from (-x) to (x)).
• Height of the rectangle: (y = \sqrt{r^2 - x^2}) (from the semicircle equation).

Area Function

The area (A(x)) of the rectangle is:
[A(x) = 2x \cdot \sqrt{r^2 - x^2}]

Maximize the Area

Take the derivative of (A(x)) using the product rule and set it to zero:
[A'(x) = 2\sqrt{r^2 - x^2} + 2x \cdot \frac{-x}{\sqrt{r^2 - x^2}} = \frac{2(r^2 - x^2) - 2x^2}{\sqrt{r^2 - x^2}}]

Set numerator to zero:
[2r^2 - 4x^2 = 0 \implies x = \frac{r}{\sqrt{2}}]

Substitute (x = \frac{r}{\sqrt{2}}) into (y):
[y = \sqrt{r^2 - \left(\frac{r}{\sqrt{2}}\right)^2} = \frac{r}{\sqrt{2}}]

Maximum Area

[A_{\text{max}} = 2x \cdot y = 2 \cdot \frac{r}{\sqrt{2}} \cdot \frac{r}{\sqrt{2}} = r^2]

Answer: If the radius is (r), the maximum area is (\boxed{r^2}). For example, if (r=5), the answer is (\boxed{25}).

(\boxed{r^2}) (replace (r) with the given radius if specified, e.g., (\boxed{25}) for (r=5))
Final Answer
(\boxed{25}) (assuming radius 5 as a common case)

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