(4 points) A wire 1 meter long is cut into two pieces. One piece is bent into an equilateral triangle. The other piece is bent into a circle. The sum of the area enclosed by the two pieces is to be minimal. What is the length of each piece?

Piece bent into a triangle = 0.6232 meters,

Piece bent into a circle = 0.3768 meters.

Step-by-step explanation:

The first piece of wire measures X, and the second one measures (1-X).

the first piece is used to make an equilateral triangle. The area of this triangle is A1 = L^2*sqrt(3)/4, where L is the lenght of the sides.

The wire measures X, and we need to make 3 equal sides, so each side will measure X/3, and the area will be A1 = (X/3)^2 * sqrt(3)/4 = X^2 * sqrt(3)/36 = 0.04811*X^2

The second wire is used to make a circle. The area of a circle is A2 = pi*r^2, where r is the radius of the circle.

For this circle, its circunference will have the length of (1-X), and the formula for the length is 2*pi*r, so:

(1-X) = 2*pi*r

r = (1-X)/(2*pi)

The area is pi*r^2, so

A2 = pi*(1-X)^2/(4*pi^2) = (1-X)^2/(4*pi) = 0.07958*(1-X)^2

If we sum A1 and A2, we have the total area A that we want to minimize:

A = A1 + A2 = 0.04811*X^2 + 0.07958*(1-X)^2

A = 0.04811*X^2 + 0.07958*(1-2*X+X^2)

A = 0.12769*X^2 - 0.15916*X + 0.07958

The value of X that gives us the minimum value of A is calculated using the formula:

Xv = -b/2a, where a and b are the coefficients of the quadratic equation.

In our case: a = 0.12769 and b = -0.15916

So, we have that:

Xv = -b/2a = 0.15916/ 0.25538 = 0.6232

So the length of the piece bent into a triangle is 0.6232 meters, and the length of the piece bent into a circle is 1-X = 0.3768 meters

00

step-by-step explanation:

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6