Question

A wire of length 10 m is divided into two pieces and each piece is bent into a square. How should this be done in order to minimize the sum of the areas of the two squares

Answer 1

A(x) = (x/4)^2 + ((10 - x)/4)^2

A'(x) = 2(x/4)(1/4) + 2((10 - x)/4)(-1/4)

x/8 + (x - 10)/8 = 0

x + x - 10 = 0

2x = 10, so x = 5

Cut the 10-meter wire into two 5-meter pieces. The area of each square is (5/4)^2 = 25/16 = 1.5625 square meters, so the combined area for both squares is 25/8, or 3.125 square meters (1.5625 square meters per square).