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?

Question

Ezra Adams · Accepted Answer

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

Ezra Adams · Answer

undefined To maximize the total area of the square and the equilateral triangle formed from the 6-inch wire, we need to use the wire efficiently.

1. **Forming the Square:** Since a square has four equal sides, divide the wire into four equal parts, each measuring 6/4 = 1.5 inches. Use these four parts to form the four sides of the square. The area of the square is then side^2, so each side is 1.5 inches, making the area 1.5^2 = 2.25 square inches.

2. **Forming the Equilateral Triangle:** An equilateral triangle also has three equal sides. Divide the remaining wire (6 - 4*1.5 = 0 inches) into three equal parts, each measuring 0 inches. Use these three parts to form the three sides of the equilateral triangle. The area of an equilateral triangle is (side^2 * sqrt(3)) / 4, where side is the length of each side. Since each side is 0 inches, the area of the equilateral triangle is 0 square inches.

Therefore, to maximize the total area contained by the square and the equilateral triangle, we would form only the square, as it has a positive area.