Problem 6
Suppose a rectangle is inscribed in a 30, 60, 90 degree triangle with one side of the rectangle along the hypotenuse of the triangle. If the length of the hypotenuse is h, what is the maximum possible area of the rectangle in terms of h?
[Problem
submitted by Vin Lee, LACC Associate Professor of
Mathematics.]
Solution for Problem 6:

Notice that
the three smaller triangles are also 30, 60, 90-degree triangles. In all such
triangles the ratio of the lengths of the sides is
. Using this ratio we
find
. So, ![]()
The area of the
rectangle is ![]()
![]()
.
This is a quadratic
equation of the form
. Since
is negative the graph
of A versus w is a downward opening parabola.
The maximum possible area is at the vertex whose w coordinate is
. So,
, and the maximum
area is
.