A piece of wire 29 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? 29 m (b) How much wire should be used for the square in order to minimize the total area?

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

Differentiate both sides with respect to d.

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

unbalanced force

explanation: