A piece of wire 23 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)(a) How much wire should be used for the square in order to maximize the total area?m(b) How much wire should be used for the square in order to minimize the total area?m

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;  link

what are the options? ?

step-by-step explanation:

the first one

step-by-step explanation:

just id the test.