Largest Rectangle inside a Circle

M2007U-E2016-A2017-S2017-M2021_SMT-20160929x02

Let's draw a circle with a radius of r, and install a rectangle inside it.

From the video, the 2 angles α and β have a total of 180 degrees.

In this article, we are going to prove that the largest rectangle it can have inside is actually a square.

---

by using the cosine rule, we can know that

h² = 2r² - 2r²cosα

and

w² = 2r² - 2r²cosβ

w² = 2r² - 2r²cos(180° - α)

w² = 2r² - 2r²(cos180° cosα - sin180°sinα)

w² = 2r² - 2r²((-1) cosα - (0)sinα)

w² = 2r² + 2r²(cosα)

therefore, to get the area of the rectangle, we can have

h²w² = (2r² - 2r²cosα)(2r² + 2r²cosα)

h²w² = (2r²)² - (2r²cosα)²

h²w² = 4r⁴ - 4r⁴cos²α

hw = (4r⁴ - 4r⁴cos²α)^(1/2) = A

to make sure that the area has a maximum value, we just need to differentiate this equation twice and see if the result is negative or not.

dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ d(4r⁴ - 4r⁴cos²α)/dα

dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ d(-4r⁴(cosα)²)/dα

dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ (-8r⁴cosα) ⋅ d(cosα)/dα

dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ (-8r⁴cosα) ⋅ (-1)sinα

dA/dα = [(-8r⁴cosα) ⋅ (-1)sinα] / [2 ⋅ (4r⁴ - 4r⁴cos²α)^(1/2)]

dA/dα = [8r⁴cosα ⋅ sinα] / [2 ⋅ ((4r⁴)(1 - cos²α))^(1/2)]

dA/dα = [8r⁴cosα ⋅ sinα] / [2 ⋅ (4r⁴sin²α)^(1/2)]

dA/dα = [8r⁴cosα ⋅ sinα] / [2 ⋅ 2r²sinα]

dA/dα = 2r²cosα

d²A/dα² = d(2r²cosα) /dα

d²A/dα² = 2r²(1) ⋅ d(cosα) /dα

d²A/dα² = 2r²(1) ⋅ (-1)sinα

d²A/dα² = -2r²sinα

since we are dealing with 0° ≤ α ≤ 180°, then sinα > 0, then -2r²sinα < 0, therefore d²A/dα² < 0, meaning the Area of the rectangle has a valid maximum value.

If we plot a graph for A and α, when A reaches the maximum value, dA/dα = 0

from this we can know that

when

dA/dα = 0 = 2r²cosα

cosα = 0

α = 90°

and when α = 90°, then β = 90°

the whole shape inside the circle is a square.

---

Do you know ?

This proof is inspired by the main task of filling up a circular region in Minecraft by using the command /fill in order to achieve the maximum efficiency