A piece of wire 7 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

Question

A. How much wire should be used for the square in order to maximize the total area?
B. How much wire should be used for the square in order to minimize the total area?

Answer 1

let x = side of triangle
let s = side of square
then
7 = 3 x + 4 s so s = (1/4)[7-3x]
A = (1/4)x^2sqrt 3 + s^2
A = (1/4)x^2 sqrt 3 +(1/16)[49-42x+9x^2]
dA/dx= (x/2)sqrt 3 +(1/16)[18x-42]
set to zero for max and min value of x
for which is max
d^2A/dx^2 = (1/2)sqrt3 + (9/8)x
if positive, min
if negative, max

Remember x is the triangle side, s = (1/4)(7-3x)