What is the area of the largest rectangle that can be inscribed in a semi-circle with a radius of 5?

Finding the Area of the Largest Inscribed Rectangle in a Semi-Circle

To determine the area of the largest rectangle inscribed in a semi-circle with a radius of 5, we can use some geometry and calculus.

1. Understanding the Geometry

Consider a semi-circle with center at the origin (0,0) in the Cartesian plane, extending from -5 to 5 on the x-axis. The equation of the semi-circle is:

y = √(25 - x²)

The rectangle will have its base on the x-axis and its top corners on the semi-circle. Let’s say the rectangle extends from (-x, 0) to (x, 0) and its height is determined by the semi-circle at that x value, which is √(25 – x²).

2. Area of the Rectangle

The area (A) of the rectangle can be expressed as:

A = width × height

Here, the width is 2x (from -x to x), and the height is √(25 – x²). Therefore:

A = 2x * √(25 - x²)

3. Maximizing the Area

To find the value of x that maximizes the area, we can differentiate the area with respect to x:

Let A = 2x * √(25 - x²)

Using the product rule, we differentiate A:

dA/dx = 2√(25 - x²) + 2x * (1/2)(25 - x²)^(-1/2)(-2x) = 2√(25 - x²) - (2x²/(√(25 - x²)))

Setting this derivative equal to zero gives us the critical points:

2√(25 - x²) - (2x²/(√(25 - x²))) = 0

4. Solving the Equation

Rearranging the equation leads to:

2√(25 - x²) = (2x²/(√(25 - x²)))

Squaring both sides to eliminate the square root (and assuming x > 0):

4(25 - x²) = 4x²

Simplifying gives:

100 - 4x² = 4x²
100 = 8x²
x² = 12.5
x = √12.5 = 5/√2 = 5√2/2

5. Calculating the Maximum Area

Now that we have x, we can find the corresponding height:

height = √(25 - (5√2/2)²) = √(25 - 12.5) = √12.5

Then, the area of the rectangle can be calculated as:

A = 2(5√2/2) * √(12.5) = 5√2 * √12.5

Consolidating the square roots:

A = 5√(25) = 5 * 5 = 25

Conclusion

Thus, the area of the largest rectangle that can be inscribed in a semi-circle of radius 5 is 25 square units.

Leave a Comment