A piece of wire 11 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

(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?

If the triangle has side x, then the area

a = ((11-3x)/4)^2 + √3/4 x^2

That's just a parabola. Its vertex will be the minimum area. For maximum, check to see which is greater: a(0) or a(11).

To determine the amount of wire that should be used for the square to maximize and minimize the total area, we need to set up the problem mathematically and find the derivative of the area equation. Let's proceed step by step:

(a) To maximize the total area, we need to find the length of a side of the square that will yield the maximum possible area.

Let's assume the length of wire used for the square is 'x'. Therefore, the length of wire used for the equilateral triangle would be (11 - x) since the total length of wire is 11 m.

For a square, the perimeter is four times the length of a side, so the length of a side of the square is (x/4).

The area of a square is given by the equation: A_square = (side)^2 = (x/4)^2 = (x^2)/16.

The area of an equilateral triangle is given by the equation: A_triangle = sqrt(3)*(side^2)/4 = (sqrt(3)*(x^2))/48.

The total area (A_total) is the sum of the areas of the square and the equilateral triangle: A_total = (x^2)/16 + (sqrt(3)*(x^2))/48.

To maximize the total area, we need to find the value of 'x' that maximizes the function A_total.

To do that, we differentiate A_total with respect to 'x' and find the critical points, which will give us the potential maximum or minimum points.

dA_total/dx = (2x)/16 + (sqrt(3)*(2x))/48.
Simplifying further, we get: dA_total/dx = x/8 + (sqrt(3)*x)/48.

Setting dA_total/dx equal to zero:
x/8 + (sqrt(3)*x)/48 = 0.

Solving this equation for 'x', we find: x = 0 or x = 0.

However, since x = 0 doesn't make sense in this context, we discard that solution.

Thus, the only critical point is x = 0.

To determine whether this critical point is a maximum or minimum, we take the second derivative:

d^2A_total/dx^2 = 1/8 + (sqrt(3)/48) > 0.

Since the second derivative is positive, we conclude that the critical point x = 0 is a minimum point.

Therefore, to maximize the total area, we need to use the entire 11 m of wire for the square. So, the answer to part (a) is 11 m.

(b) To minimize the total area, we need to determine the length of a side of the square that will yield the smallest possible area.

Since we have already solved for the critical point in part (a), we know that the minimum occurs at x = 0.

Therefore, the amount of wire used for the square to minimize the total area is 0 m.

So, the answer to part (b) is 0 m.

In summary:
(a) The wire used for the square to maximize the total area should be 11 m.
(b) The wire used for the square to minimize the total area should be 0 m.