A piece of wire 40 m long is cut into two pieces. One piece is bent
into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is a
maximum=
minimum=
Find the length of the wire used
If the sides are s and , respectively,
4s+3t = 40
a = √3/4t^2 + s^2
now express a as a function of s (or t) only, and find its extrema (where a'=0)
(sqrt(3)/4)t^2+((40-3t)/4)^2
81
To find the cut for the maximum area, we need to consider the formulas for the area of a square and an equilateral triangle.
1. Maximum area:
Let x represent the length of the wire used for the square, so the remaining (40 - x) is used for the equilateral triangle. The perimeter of the square is 4 times the length of one side, which is equal to x. Thus, the side length of the square is x/4.
The area of the square is given by A_square = (side length)^2 = (x/4)^2 = x^2 / 16.
The perimeter of the equilateral triangle is 3 times the length of one side, which is equal to (40 - x). Thus, the side length of the equilateral triangle is (40 - x) / 3.
The area of an equilateral triangle is given by A_triangle = √3/4 * (side length)^2 = √3/4 * ((40 - x) / 3)^2 = √3(40 - x)^2 / 36.
The total area, A_total, is the sum of the area of the square and the area of the triangle: A_total = x^2 / 16 + √3(40 - x)^2 / 36.
To find the maximum area, we can take the derivative of A_total with respect to x, set it equal to zero, and solve for x. Then substitute x back into the equation to find the area.
2. Minimum area:
To find the cut for the minimum area, we can follow a similar process. The only difference is that this time, we want to find the minimum area, so we need to find the minimum value of A_total. Again, we can find this by taking the derivative of A_total with respect to x, setting it equal to zero, and solving for x.
3. Finding the length of the wire used:
Given the two pieces of wire, x and (40 - x), the length of the wire used is simply the sum of these two: Length of wire used = x + (40 - x) = 40.
To summarize:
- To find the cut for the maximum area, differentiate A_total with respect to x, set it equal to zero, and solve for x.
- To find the cut for the minimum area, differentiate A_total with respect to x, set it equal to zero, and solve for x.
- The length of the wire used is always 40, as it is stated in the question.