A square and a equilateral triangle are to be formed out of the same piece of wire. The wire is 6 inches long. How do you maximize the total area the square and the triangle contain?

1 Answer
May 13, 2016

# (4sqrt(3)L)/(9+4sqrt(3))# for the square
# (9L)/(9+4sqrt(3))# for the equilateral triangle
Here #L = 6#

Explanation:

Let be #L = s + t# the total length as the addition of #s# the length used by the square and #t# the length used by the triangle.
The square area is #a_s = (s/4)^2# and the equilateral triangle area is given by #a_t = (t/6)sqrt((t/3)^2-(t/6)^2) = t^2/(12 sqrt(3))#
the total area is then #a = a_s+a_t = s^2/16+t^2/(12sqrt(3))#
but #t = L-s# then #a = (L-s)^2/(12sqrt(3))+s^2/16#
The area critical point is determined doing #(da)/(ds) = 0# and obtaining #s = (4sqrt(3)L)/(9+4sqrt(3))# and also #t = (9L)/(9+4sqrt(3))#