Respuesta :
A) The amount of wire that should be used for the square in order to maximize the total area is; All the wire of 23 m
B) The amount of wire that should be used for the square in order to minimize the total area is; 10m of wire
Let's say the perimeter one side of the square is x meters.
Then, the length of the side of the square will be = x/4 meters
Area of the square will be;
A₁ = x/4 • x/4 = x²/16 m²
Now, since x is used for the square, the remaining will be 23 - x and thus, if one side of the equilateral triangle is given as "a", hence;
a = (23 − x)/3 m
Area of equilateral triangle will be;
A₂ = (√3)a²/4
A₂ =((√3)/36) • (23 − x)² m²
Hence the total area is;
A = A₁ + A₂
A = x²/16 + ((√3)/36) • (23 − x)²)
Now let's find the first and second derivatives of this total area;
dA/dx = x/8 − ((√3)/18)(23 - x)
and;
d²A/dx² = 1/8 + ((√3)/18) = 0.22
Thus, d²A/dx² > 0
Therefore, the extremum is at dA/dx = 0
Thus,
x/8 − ((√3)/18)(23 - x) = 0
x/8 = ((√3)/18)(23 - x)
x = ((8√3)/18)(23 - x)
x = 0.7698(23-x)
x = 17.7054 - 0.7698x
x + 0.7698x = 17.7054
1.7698x = 17.7054
x = 17.7054/1.7698 ≈ 10m
Since d²A/dx² > 0, this is a minimum.
Thus, the minimum area is obtained when 10m of wire is used for the square.
B) For a given perimeter, a square normally has more area than a triangle. Therefore, to get the largest or maximum area we need to use all of the wire to construct the square. Which is 23m
Read more about minimization operations at; https://brainly.com/question/15314150
