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Thursday, January 21, 2010

Consider an equilateral triangle with sides of length 20. Find the area of the largest rectangle that can be inscribed in the triangle.

Chris said...: Amax = 50√3. This is exactly 1/4 of the triangle's area.; January 21, 2010 10:32 PM
Chris said...: Let 2c be length of a side of the triangle.
Let the triangle have one edge along the x-axis, and be centred about the y-axis.

The equation of the right hand corner of the rectangle is
y = (c-x)√3

Area of rectangle = A = 2xy = 2√3x(c-x)
= 2√3(xc - x²)

dA/dx = 2√3(c - 2x) = 0 => x = c/2

Amax =2√3c(c-c/2)/2 = √3c²/2

c = 10 => Amax = 50√3

The area of the triangle = 200√3; January 21, 2010 11:05 PM
Zaux said...: Chris,
you are once again correct; January 22, 2010 5:58 AM
Anonymous said...: Trigonometric Approach:

if the triangle is pointing up, insert the rectangle with its base on the triangle's base and its top corners at the mid points of the triangle's sides.

Note: Using degrees in trig calculations

The height of the rectangle is:
(20/2)*sin(60) = 5*√3

The base of the rectangel is:
20-2*(10*cos(60)) = 10

Area of Rectangle:
b*h = 10*(5*√3) = 50*√3 = 86.60; January 22, 2010 7:32 PM
Chris said...: Anonymous. How did you know that the corners of the largest area rectangle would be mid-way along the sloping sides of the triangle? I don't think that would be obvious to most people. It isn't to me.; January 22, 2010 9:14 PM