Question

A piece of wire 26 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? 26 m (b) how much wire should be used for the square in order to minimize the total area?

Answer 1

To maximize the total area, we need to determine how much wire should be used for the square. The formula for the perimeter of a square is P = 4s, where s is the length of each side. In this case, the length of each side of the square is (x/4) meters.

Step-by-step explanation:

To maximize the total area, we need to determine how much wire should be used for the square. Let's assume that x meters of wire are used for the square, then the remaining length of the wire (26 - x) will be used for the equilateral triangle.

The formula for the perimeter of a square is P = 4s, where s is the length of each side. In this case, the length of each side of the square is (x/4) meters.

The formula for the perimeter of an equilateral triangle is P = 3s, where s is the length of each side. In this case, the length of each side of the equilateral triangle is ((26 - x)/3) meters.

To maximize the total area, we need to find the values of x that will maximize the area of the square and the area of the equilateral triangle:

A_square = (x/4)^2

A_triangle = (sqrt(3)/4) * ((26 - x)/3)^2

The total area, A_total, is given by:

A_total = A_square + A_triangle

Using calculus, we can find the value of x that maximizes A_total. By taking the derivative of A_total with respect to x and setting it equal to zero, we can find the critical point:

dA_total/dx = 0

Solving this equation will give us the value of x that maximizes A_total.

Answer 2

a) All the 26m should be used to bent into a square to maximize the total area (= 42.25 m2) b) 1.4136m should be bent into a square and the rest (= 24.586m) should be bent into an equilateral triangle to minimize the total area (= 29.0835m2)