Question

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?

Answer 1

`(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))`