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 much of the wire should go to the square to minimize the total area enclosed by both figures?
let the side of the square be x
let the side of the triangle be y
4x + 3y = 40
x = (40-3y)/4
height of triangle from the 30-60-90 triangle ratio = √3/2 y
area of triangle = (1/2)(y)(√3/2 y)
= (√3/4)y^2
area of square = [(40-3y)/4]^2
= (1600 - 240y + 9y^2)/16
= 100 - 15y + (9/16)y^2
A = 100 - 15y + (9/16)y^2 + (√3/4)y^2
dA/dy =-15 + 9/8 y + 2√3/4 y
= 0 for a max/min
√3/2 y + 9/8 y = 15
4√3 y + 9y = 135
y = 135/(4√3+9) = appr 8.476
3y = 25.43 m
then 4x = 14.573 m
appr 14.57 m should go for the square
check my arithmetic and algebra
To minimize the total area enclosed by both figures, we need to determine the amount of wire that should go to the square.
Let's assume the length of the wire used for the square is x meters.
The remaining length of the wire will be (40 - x) meters, which will be used for the equilateral triangle.
Now, let's find the perimeter of the square.
Since a square has four equal sides, each side will have a length of x/4.
Therefore, the perimeter of the square is 4 * (x/4) = x.
Next, let's find the side length of the equilateral triangle.
Since an equilateral triangle has three equal sides, each side will have a length of (40 - x)/3.
The perimeter of the equilateral triangle is 3 * [(40 - x)/3] = 40 - x.
To minimize the total area enclosed by both figures, we need to minimize the sum of their areas.
The area of a square is given by A(square) = (side length)^2 = (x/4)^2 = x^2/16.
The area of an equilateral triangle is given by A(triangle) = (sqrt(3)/4) * (side length)^2 = (sqrt(3)/4) * [(40 - x)/3]^2.
The total area, A(total), is given by A(square) + A(triangle).
So, A(total) = x^2/16 + (sqrt(3)/4) * [(40 - x)/3]^2.
To minimize A(total), we need to find the value of x that minimizes it.
We can take the derivative of A(total) with respect to x and set it equal to zero to find the critical point.
d/dx (A(total)) = (2x)/16 - (sqrt(3)/4) * 2 * (40 - x)/3 = 0.
Simplifying, we get:
x/8 - (2sqrt(3)/12) * (40 - x)/3 = 0.
Multiplying through by 24, we get:
3x - 2sqrt(3) * (40 - x) = 0.
Expanding and simplifying, we get:
3x - 2sqrt(3) * 40 + 2sqrt(3)x = 0.
Grouping like terms, we get:
5x + 2sqrt(3)x = 2sqrt(3) * 40.
Combining the terms, we get:
7x = 80sqrt(3).
Dividing by 7, we get:
x = 80sqrt(3)/7.
Therefore, approximately 7.97 meters of the wire should go to the square in order to minimize the total area enclosed by both figures.
To minimize the total area enclosed by both figures, we need to optimize the lengths of the wire allocated for the square and the equilateral triangle.
Let's consider that a length 'x' is used for the side of the square. Since a square has four equal sides, the total length of wire used for the square is 4x.
Now, let's consider that a length 'y' is used for each side of the equilateral triangle. Since an equilateral triangle has three equal sides, the total length of wire used for the triangle is 3y.
According to the problem statement, the total length of the wire is 40 m. So, we can write the equation:
4x + 3y = 40 ...(Equation 1)
To find the values of 'x' and 'y' that minimize the total area enclosed by both figures, we need to express the area in terms of 'x' and 'y'.
The area of a square is given by A_square = x^2.
The area of an equilateral triangle is given by A_triangle = (√3 / 4) * y^2.
The total area enclosed by both figures is the sum of the areas of the square and the triangle:
Total Area (A_total) = A_square + A_triangle = x^2 + (√3 / 4) * y^2
To minimize the total area, we can take the derivative of A_total with respect to 'x' and 'y' and set them equal to zero.
dA_total / dx = 2x = 0 ...(Equation 2)
dA_total / dy = [(√3 / 4) * 2y] = (√3 / 2) * y = 0 ...(Equation 3)
From Equation 2, we can find that x = 0, which is not a valid solution since the length cannot be zero.
From Equation 3, we can find that y = 0, which is also not a valid solution.
Since neither x = 0 nor y = 0 is valid, we need to consider the boundary conditions. By substituting the value of x from Equation 1 into the Total Area equation, we get:
A_total = (40 - 3y)/4 * (40 - 3y)/4 + (√3 / 4) * y^2
Expanding and simplifying the equation, we can rewrite it as:
A_total = (1600 - 240y + 9y^2)/16 + (√3 / 4) * y^2
To find the minimum value of A_total, we need to find the values of 'y' that satisfy:
dA_total / dy = 0
Taking the derivative of A_total with respect to 'y' and setting it equal to zero:
dA_total / dy = (-240 + 18y)/16 + (√3 / 4) * 2y = 0
Simplifying the equation:
-240 + 18y + (√3 / 2) * 8y = 0
Combining like terms:
18y + 4√3y = 240
Solving for 'y':
22.97y = 240
y ≈ 10.45
Now, substitute the obtained value of 'y' back into Equation 1 to find the corresponding value of 'x':
4x + 3(10.45) = 40
4x = 40 - 31.35
4x ≈ 8.65
x ≈ 2.16
Therefore, to minimize the total area enclosed by both figures, approximately 2.16 meters of wire should go to the square, and approximately 10.45 meters of wire should go to the equilateral triangle.