A 25 inch copper wire will be used to form two squares. This is done by cutting the 25 inch
wire into two parts. Minimize the possible sum of the two areas of the square
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Expecting the two squares to be equal .... anyway
let side of the first square be x
let the side of the 2nd square be y
4x + 4y = 25
y = (25-4x)/4
total area = x^2 + y^2
= x^2 + ((25 - 4x)/4)^2
= x^2 + (625 - 200x + 16x^2 )/16
d(total area)/dx = 2x - 25/2 + 2x
= 4x - 25/2
= 0 for a min of area
4x = 25/2
x = 25/8
y = (25 - 4(25/8)) / 4 = 25/8
sure enough, since x = y, the two squares are equal
with each square equal to (25/8)^2 square units
To minimize the sum of the two areas of the square, we need to maximize the size of each square.
Let's denote the length of one side of each square as x. Since the wire is cut into two parts, we can represent the total length of wire used as:
Perimeter of Square 1 + Perimeter of Square 2 = 2 * (Length of one side of Square 1 + Length of one side of Square 2)
Since each square has four equal sides, we can rewrite this equation as:
4x + 4x = 2 * 25
Simplifying this equation gives us:
8x = 50
Dividing both sides of the equation by 8, we get:
x = 50/8 = 6.25
Since side length cannot be negative, we discard the negative solution. So, the length of one side of each square is 6.25 inches.
To find the sum of the two areas of the square, we square the length of one side and then multiply by 2:
Sum of Areas = 2 * (x * x) = 2 * (6.25 * 6.25) = 2 * 39.06 = 78.12 square inches
Therefore, by cutting the 25-inch wire into two parts and forming two squares, the minimum possible sum of the two areas of the squares is 78.12 square inches.