Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle.
base=
height=

Let x represent the ratio of the rectangle base to the triangle side length. Then the height of the small triangle above the rectangle will be x times the height of the equilateral triangle. Then the height of the rectangle is (1-x) times the height of the equilateral triangle. The rectangle's area will be ...

A = bh

A = (xL)(1-x)(L·√3/2) = (L²√3/2)(x)(1-x)

This graphs as parabola opening downward with x-intercepts at x=0 and x=1. The vertex is on the line of symmetry, halfway between these zeros, at x = 1/2.

The base of the rectangle is L/2.

The height of the rectangle is L√3/2.

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The general solution to this sort of problem is that one side of the rectangle is the midsegment of the triangle.

Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle. base= height=

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Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle.
base=
height=