A five feet piece of wire is cut into two pieces. One Piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire be cut so that the total area enclosed by both is minimum.
let each side of the equilateral triangle be 2x
let each side of the square be y
then 6x + 4y = 5
y = (5-6x)/4
Area of triangle:
draw in the altiude, so we have a right angled triangle with hypotenuse 2x and base 1x, by Pythagorus the height is √3x (now do you see why I defined the side as 2x and not x ? )
Area of triangle = (1/2)(2x)√3x = √3x^2
area of square = y^2
Area = √3x^2 + y^2
= √3x^2 + ((5-6x)/4 )^2
= √3x^2 + (1/16) (25 - 60x + 36x^2)
= √3x^2 + 25/16 - (15/4)x + (9/4)x^2
d(Area)/dx = 2√3x - 15/4 + (9/2)x
= 0 for a min of Area
x(2√3 + 4.5) = 3.75
x = 3.75/(2√3+4.5) = appr .4709
so each side of the triangle is 2x = .9417 ft
and each side of the square = (5-6x)/4 = .5437 ft
check my arithmetic, I should have written it out on paper first.
To find the location where the wire should be cut in order to minimize the total area enclosed by both the square and the equilateral triangle, we can approach it mathematically step by step.
Let's assume the length of the wire used for the square is "x" feet. Therefore, the length of wire used for the equilateral triangle will be (5 - x) feet.
Now let's calculate the areas of the square and the equilateral triangle:
1. Area of the Square:
The perimeter of a square is equal to four times the length of its side (in this case, x). Therefore, the length of each side of the square will be x/4.
The area of the square is given by the formula: Area = side^2. So, the area of the square will be (x/4)^2.
2. Area of the Equilateral Triangle:
The perimeter of an equilateral triangle is equal to three times the length of its side (in this case, 5 - x). Therefore, the length of each side of the equilateral triangle will be (5 - x)/3.
The area of the equilateral triangle is given by the formula: Area = (sqrt(3)/4) * side^2. So, the area of the equilateral triangle will be (sqrt(3)/4) * ((5 - x)/3)^2.
Now, to find the total area enclosed by both shapes, we add the areas of the square and the equilateral triangle:
Total Area = (x/4)^2 + (sqrt(3)/4) * ((5 - x)/3)^2
To find the minimum total area, we need to find the value of "x" that minimizes the equation.
We can use calculus to find the minimum of the function. Take the derivative of the function with respect to "x" and set it equal to zero:
d(Total Area) / dx = 0
By solving this equation, we can find the value of "x" that minimizes the total area.
After finding the value of "x," we can determine the position at which the wire should be cut to achieve the desired minimum total area.