A wire of length 4 m is cut in two parts and the first part is bent in the shape of a square, while the second part is bent in the shape of an equilateral triangle. For which lengths of these parts the total area enclosed by the square and the triangle is: a. minimized? b. maximized?

if the square has side s and the triangle has side t (both in cm), then

4s+3t = 400
a = s^2 + √3/4 t^2

Now just express t or s in terms of the other, and you have a quadratic for the area. The vertex of the parabola will give the minimum area.

The maximum will be realized when all of the wire is used for one shape.

To find the lengths that minimize and maximize the total area enclosed by the square and the triangle, let's break down the problem and analyze it step by step.

Let's assume the length of the first part, which is bent into a square, is x meters. Therefore, the length of the second part, bent into an equilateral triangle, would be 4 - x meters.

a. To minimize the total area, we need to determine the lengths of the parts that result in the smallest possible area for the square and the triangle combined.

The area of a square is given by the formula A_square = side², where side is the length of any side of the square.

The area of an equilateral triangle is given by the formula A_triangle = (side² * √3) / 4, where side is the length of any side of the equilateral triangle.

Therefore, the total area, A_total, is equal to A_square + A_triangle.

A_total = (x²) + [(4 - x)² * √3 / 4]

Simplifying the equation:
A_total = x² + (16 - 8x + x²) * √3 / 4
A_total = x² + (8 + x² - 8x) * √3 / 4
A_total = (2x² - 8x + 8) * √3 / 4
A_total = (√3 / 2)x² - 2√3x + 2√3

Now, to find the minimum area, we can use calculus by taking the derivative of A_total with respect to x, setting it equal to zero, and solving for x.

dA_total/dx = (√3 / 2)(2x) - 2√3 = 0
√3x - √3 = 0
x = 1

Therefore, to minimize the total area, the length of the first part (the square) should be 1 meter, and the length of the second part (the equilateral triangle) should be 4 - 1 = 3 meters.

b. To maximize the total area, we can use the same equation for A_total and find the value of x that yields the largest possible area.

A_total = (√3 / 2)x² - 2√3x + 2√3

Again, we'll take the derivative of A_total with respect to x, set it equal to zero, and solve for x.

dA_total/dx = (√3 / 2)(2x) - 2√3 = 0
√3x - √3 = 0
x = 1

Similar to part (a), we find that the length of the first part (the square) should be 1 meter, and the length of the second part (the equilateral triangle) should be 4 - 1 = 3 meters to maximize the total area.

In both cases, the lengths that minimize and maximize the total area enclosed by the square and the triangle are 1 meter and 3 meters respectively.