A piece of wire 14 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?
= 14 m
(b) How much wire should be used for the square in order to minimize the total area?
= ????
let each side of the square be x m
let each side of the equilateral triangle be 2x
(that way, the height is √3y, from the ratio of the 30-60-90° triangle)
a) for a max area, you are right, all should be used for the square
b) 4x + 6y = 14
2x + 3y = 7
x = (7-3y)/4 OR y = (7-2x)/3
area = x^2 + (1/2)(2y)(√3y)
= x^2 + √3 y^2
= x^2 + √3 ((7-2x)/3)^2
= x^2 + (√3/9)(49 - 28x + 4x^2)
d(area)/dx = 2x + (√3/9)(-28 + 8x) = 0 for a max of area
2x = √3/9(28 - 8x)
18x = 28√3 - 8√3x
x(18 + 8√3) = 28√3
x = 28√3/(18+8√3) = appr 1.522
need 4 x's for the square, so 6.09 m for the square, leaving 7.9 m for the triangle for a minimum total area.
check my arithmetic, should have written it out first.
To find out how much wire should be used for the square in order to maximize the total area, we can start by setting up the problem and expressing the quantities involved mathematically.
Let's assume that x meters of wire is used for the square. This means that the remaining (14 - x) meters of wire is used for the equilateral triangle.
(a) To maximize the total area, we need to maximize the area of the square. The area of a square is given by A = s^2, where s is the length of one side of the square.
In this case, the perimeter of the square is equal to the total amount of wire used for the square: 4s = x (since the square has four equal sides).
Therefore, the side length of the square can be expressed as s = x/4.
Now, substituting the value of s in terms of x into the area equation, we get A = (x/4)^2 = x^2/16.
To maximize the total area, we need to maximize this expression, which means finding the maximum value of x^2/16.
Since x^2 is always positive, x^2/16 will be maximized when x^2 is maximized. This occurs when x is at its maximum value. In this case, x is limited to a maximum length of 14 meters (the length of the wire available).
Therefore, in order to maximize the total area, the entire length of wire (14 meters) should be used for the square, and the remaining 0 meters would be used for the equilateral triangle.
(b) To minimize the total area, we need to minimize the area of the square. Similar to the previous case, the area of the square is given by A = x^2/16.
To minimize the total area, we need to minimize this expression. Since x^2 is always positive, the minimum value of x^2/16 will occur when x^2 is minimized. This occurs when x is at its minimum value.
In this case, the minimum value of x is 0 meters since the wire hasn't been used for the square at all.
Therefore, in order to minimize the total area, 0 meters of wire should be used for the square, and the entire length of wire (14 meters) would be used for the equilateral triangle.