Prove that the rectangle of maximum area inscribed in a circle is a square.

To prove that the rectangle of maximum area inscribed in a circle is a square, we can follow these steps:

Step 1: Set up the problem
Let the equation of the circle be:

x² + y² = r²

where r is the radius of the circle.

We are to inscribe a rectangle inside this circle such that the rectangle has the maximum area. The rectangle's vertices lie on the circle.

Step 2: Define the coordinates of the rectangle's vertices
Assume that one vertex of the rectangle is at (x, y), and since the rectangle is inscribed, the opposite vertex will be at (-x, -y). The other two vertices will be at (x, -y) and (-x, y).

Step 3: Express the area of the rectangle
The area of the rectangle is given by:

A = length × width

The length is the distance between the points (x, y) and (x, -y), which is 2y.

The width is the distance between the points (x, y) and (-x, y), which is 2x.

Thus, the area of the rectangle is:

A = 2x * 2y = 4xy

Step 4: Use the constraint from the circle equation
Since the rectangle is inscribed in the circle, the point (x, y) must satisfy the equation of the circle:

x² + y² = r²

Step 5: Maximize the area using calculus
To maximize the area A = 4xy, we can express y in terms of x using the circle's equation:

y² = r² - x² y = √(r² - x²)

Substitute this into the area formula:

A(x) = 4x * √(r² - x²)

Now, differentiate A(x) with respect to x to find the critical points:

dA/dx = 4 * [√(r² - x²) + x * (-x/√(r² - x²))]

Simplifying:

dA/dx = 4 * [√(r² - x²) - x² / √(r² - x²)]

Set dA/dx = 0 to find the critical points:

√(r² - x²) = x² / √(r² - x²)

Square both sides:

r² - x² = x⁴ / (r² - x²)

Multiply both sides by (r² - x²):

(r² - x²)² = x⁴

Expand both sides:

r⁴ - 2r²x² + x⁴ = x⁴

Simplify:

r⁴ - 2r²x² = 0

Factor the equation:

r²(r² - 2x²) = 0

Since r² ≠ 0, we have:

r² - 2x² = 0 x² = r² / 2 x = r / √2

Step 6: Find the corresponding value of y
Substitute x = r / √2 into the circle's equation to find y:

x² + y² = r² (r / √2)² + y² = r² r² / 2 + y² = r² y² = r² - r² / 2 = r² / 2 y = r / √2

Step 7: Check the area
The length and width of the rectangle are both 2x and 2y, respectively:

Length = 2x = 2r / √2 Width = 2y = 2r / √2

The area is:

A = length × width = (2r / √2) * (2r / √2) = 2r²

Step 8: Conclude that the rectangle is a square
Since the length and width are equal (2r / √2), the rectangle is a square.

Thus, we have proved that the rectangle of maximum area inscribed in a circle is a square.