a)The problem provides you the information that the piece of wire is cut into two pieces, a square and equilateral triangle.

You should come up with the substitutions for the perimeters of the equilateral triangle and perimeter of square such that:

x = perimeter of square

y = perimeter of equilateral triangle

The total length of 6 meters may be evaluated using the perimeters of equilateral triangle and the square such that:

`6 = x+y`

Notice that the length of the side of square is of `x/4 ` and the length of the side of equilateral triangle is of `y/3` .

You need to express the area of square such that:

A square = `(x/4)^2` => A square = `(x^2)/16`

You need to express the area of equilateral triangle such that:

A triangle = ((y^2)/9)*(sqrt3/4)

You need to maximize the total area, hence you should differentiate the function:

`A = (x^2)/16 + ((y^2)/9)*(sqrt3/4)`

You need to write the total area in terms of one variable, hence you may use the equation x+y= 6=>x=6-y.

`A(y) = ((6-y)^2)/16 + ((y^2)/9)*(sqrt3/4)`

You need to differentiate the function with respect to y such that:

`A'(y) = -2(6-y)/16 + (2y/9)*(sqrt3/4)`

`A'(y) = -(6-y)/8 + (y/9)*(sqrt3/2)`

You need to solve the equation A'(y) = 0 such that:

`-(6-y)/8 + (y/9)*(sqrt3/2) = 0`

`-9(6-y)+ 4sqrt3*y = 0 =gt -54 + 9y + 4sqrt3*y = 0`

`y(9+4sqrt3) = 54 =gt y = 54/(9+4sqrt3)`

Notice that the second derivative of the function is positive, hence the function reaches its minimum at `y = 54/(9+4sqrt3).`

You may find the maximum area at `x=0` and `x=6` (endpoints of wire) such that:

`A = (6^2)/16 + ((0^2)/9)*(sqrt3/4)`

`A = 36/16 =gt A = 9/4`

`A = (0^2)/16 + ((6^2)/9)*(sqrt3/4)`

`A = sqrt3`

**Hence, you should use the following lengths for the square for the total area to be maximum: `x=0 mor x=6 m` .**

b) The function A(y) reaches its minimum at `y = 54/(9+4sqrt3), ` hence, you may evaluate the length of the square such that:

`x = 6 - 54/(9+4sqrt3) =gt x = (54 + 24sqrt3 - 54)/(9+4sqrt3)`

`x = (24sqrt3)/(9+4sqrt3)`

**Hence, you should use the following lengths of wire to minimize the total area such that `x = (24sqrt3)/(9+4sqrt3) m` and `y = 54/(9+4sqrt3) m.` **

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