Question

Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius 𝑟=50.

Answer 1

[AI-generated answer — Explain Bot]

To find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius 𝑟=50, we can use calculus.

Let's consider a rectangle with sides of length 2𝑥 and 2𝑦, where 𝑥 and 𝑦 are the horizontal and vertical distances from the center of the circle to the sides of the rectangle.

The area of the rectangle can be expressed as 𝐴 = 2𝑥 * 2𝑦 = 4𝑥𝑦.

Now, since the rectangle is inscribed in a circle, the diagonal of the rectangle is equal to the diameter of the circle, which is 2𝑟.

Using the Pythagorean theorem, we have: (2𝑥)² + (2𝑦)² = (2𝑟)².

Simplifying the equation, we get: 4𝑥² + 4𝑦² = 4𝑟².

Dividing both sides of the equation by 4, we have: 𝑥² + 𝑦² = 𝑟².

Now, we need to maximize the area 𝐴 = 4𝑥𝑦, while satisfying the equation 𝑥² + 𝑦² = 𝑟².

We can solve this problem using calculus.

First, we express 𝑦 in terms of 𝑥, using the equation 𝑥² + 𝑦² = 𝑟²:

𝑦 = √(𝑟² - 𝑥²)

Now, we can substitute 𝑦 in the formula for the area 𝐴:

𝐴 = 4𝑥 * √(𝑟² - 𝑥²)

To find the maximum area, we take the derivative of 𝐴 with respect to 𝑥, set it equal to zero, and solve for 𝑥.

𝐴' = 4(√(𝑟² - 𝑥²) - 𝑥(𝑟² - 𝑥²)⁻ᵖ/²) = 0

Simplifying and solving for 𝑥:

√(𝑟² - 𝑥²) - 𝑥(𝑟² - 𝑥²)⁻ᵖ/² = 0

Squaring both sides of the equation:

(𝑟² - 𝑥²) - 𝑥²(𝑟² - 𝑥²)⁻¹ = 0

Expanding the equation:

𝑟² - 𝑥² - 𝑥²(𝑟² - 𝑥²)⁻¹ = 0

Multiplying through by (𝑟² - 𝑥²):

𝑟²(𝑟² - 𝑥²) - 𝑥² = 0

Expanding and rearranging the equation:

𝑟⁴ - 𝑟²𝑥² - 𝑥² = 0

Combining like terms:

𝑟⁴ - (𝑟² + 1)𝑥² = 0

Simplifying and solving for 𝑥²:

𝑟⁴ = (𝑟² + 1)𝑥²

𝑥² = 𝑟⁴ / (𝑟² + 1)

Taking the square root of both sides:

𝑥 = √(𝑟⁴ / (𝑟² + 1))

Substituting the value of 𝑟 = 50:

𝑥 = √(50⁴ / (50² + 1))

Simplifying:

𝑥 ≈ 34.64

Finally, using the equation 𝑦 = √(𝑟² - 𝑥²), we can find 𝑦:

𝑦 = √(50² - 34.64²)

𝑦 ≈ 41.54

Therefore, the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius 𝑟 = 50 are approximately 2𝑥 ≈ 69.28 and 2𝑦 ≈ 83.08.