Question

Greatest rectangle in a equilateral triangle by optimization

Answer 1

We’ll argue from symmetry that the maximum area rectangle has one side centered on one side of the triangle. Let’s sit the rectangle and triangle on the x axis centered around the origin.
Let’s consider the equilateral triangle of sides length two with vertices (−1,0),(1,0),(0,3–√).(−1,0),(1,0),(0,3).

Restricting ourselves to the first quadrant, we have a 30,60,90 triangle with sides the x and y axes and the line

y=3–√(1−x)y=3(1−x)

For a given xx, we have a rectangle A=2x×y=23–√x(1−x).A=2x×y=23x(1−x).

We can do calculus to maximize, or complete the square:

A/23–√=x(1−x)=x−x2=−(x2−x+14)+14=−(x−12)2+14A/23=x(1−x)=x−x2=−(x2−x+14)+14=−(x−12)2+14

That’s clearly maximized when x=12x=12giving

A=14(23–√)=3–√/2A=14(23)=3/2

We should look at that area relative to the area of the equilateral triangle of side 2,

A△=12(2)3–√=3–√A△=12(2)3=3

The maximum area inscribed rectangle is half the area of the equilateral triangle.