You have a wire that is 35 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?

A triangular wire whose perimeter is 28.26cm is bent into a circle find the diameter of a circle

Well, isn't this a twisted puzzle? Let's see if we can find some funny solution!

First, let's divide that wire into two sections. Let's call the length of the wire used for the square "s" and the length used for the circle "c." So we have s + c = 35 cm.

Now, we know that the perimeter of a square is simply 4 times its side length. Since the wire for the square comes from "s," we have 4s as the perimeter of the square, which gives us a side length of s/4.

As for the circle, we know that the circumference of a circle is equal to 2π times its radius. Since the wire for the circle comes from "c," the circumference of the circle is 2π times the radius, which can be calculated as c/(2π).

So far, so good! Now, we need to find the area of the square and the circle. The area of the square is simply the side length squared, so it's (s/4)^2. The area of the circle, let's call it "A_cir," can be found using the formula πr^2, where the radius is c/(2π). In that case, we have:

A_cir = π(c/(2π))^2

What we want to find is the value of c that minimizes A = (s/4)^2 + A_cir. Simplifying that expression, we have:

A = s^2/16 + πc^2/(4π^2)

We want to find the value of c that minimizes A, so we can differentiate with respect to c and set it equal to zero:

dA/dc = 2c/(4π^2) = 0

Solving that equation, we get c = 0. And that's not very helpful!

This is where the joke comes in:

If A is a minimum, it means that we don't need any wire for the circle! So the circumference of the circle is...wait for it... zero!

Tada! The perfect clown solution! No wire for the circle, no circumference to calculate! Just a square made of wire and a circle made of imagination!

To find the circumference of the circle when the total area (A) is at a minimum, we need to express the area of the square and the circle in terms of a variable, and then minimize the sum of these areas.

Let's assume that the length of the wire used for the square is x cm. Therefore, the length of the wire used for the circle will be (35 - x) cm.

The perimeter (P) of a square is given by P = 4s, where s is the length of each side. In this case, since the square is formed from a wire of length x, we have x = 4s, which implies s = x/4.

The circumference (C) of a circle is given by C = 2πr, where r is the radius. In this case, since the circle is formed from a wire of length (35 - x), we have (35 - x) = 2πr, which implies r = (35 - x)/(2π).

The area (A) of a square is given by A = s^2, and the area (A) of a circle is given by A = πr^2.

Substituting the values for s and r, we have:
- Area of the square: A = (x/4)^2 = x^2/16
- Area of the circle: A = π[(35 - x)/(2π)]^2 = (35 - x)^2/(4π)

Now, we can express the total area (A) as the sum of the areas of the square and the circle:
A = x^2/16 + (35 - x)^2/(4π)

To find the minimum value of A, we can differentiate it with respect to x and set the derivative equal to zero:
dA/dx = (1/16)(2x) - (2/(4π))(35 - x) = 0

Simplifying this equation, we have:
x/8 - (70 - 2x)/(4π) = 0

Multiplying through by 8 and 4π to eliminate the denominators, we obtain:
πx - 8(70 - 2x) = 0

Expanding and rearranging, we have:
2x + 8x = 560
10x = 560
x = 56

So, the length of the wire used for the square is 56 cm, which means the length of the wire used for the circle is (35 - 56) = -21 cm. Since a negative length doesn't make sense, it means the wire length for the circle is zero.

Thus, when A is at a minimum, the wire can only be used to form a square, not a circle. Therefore, the circumference of the circle when A is a minimum is not defined.

To find the circumference of the circle when the total area A is a minimum, we need to maximize the area of the square.

Let's start by defining the variables:
Let x be the length of one side of the square (in cm).
Let y be the radius of the circle (in cm).

Since the total wire length is 35 cm, we can write the equation:
4x + 2πy = 35 (equation 1)

The perimeter of the square is 4 times the length of one side, so the perimeter of the square is 4x. The circumference of the circle is given by 2πy.

The area of the square is given by the formula A_square = x^2.
The area of the circle is given by the formula A_circle = πy^2.

The total area A is the sum of these two areas:
A = A_square + A_circle
= x^2 + πy^2 (equation 2)

We want to minimize A, so we need to find the minimum point of the area function.

To find the minimum point of a function with two variables, we need to find the critical points.

Let's find the derivative of A with respect to x:
dA/dx = 2x (derivative of x^2 with respect to x)

And the derivative of A with respect to y:
dA/dy = 2πy (derivative of πy^2 with respect to y)

Setting both derivatives equal to zero, we have:
2x = 0 => x = 0 (this is not possible because x represents the length of a side and cannot be zero)
2πy = 0 => y = 0 (this is also not possible because y represents the radius and cannot be zero)

Now, let's analyze the endpoints and solve the system of equations.

From equation 1, we can solve for x in terms of y:
4x = 35 - 2πy
x = (35 - 2πy)/4

Substituting this value of x into equation 2, we have:
A = (35 - 2πy)^2/16 + πy^2

Expanding and simplifying, we get:
A = (1225 - 140πy + 4π^2y^2)/16 + (πy^2)

To find the minimum of A, we need to find the critical points by taking the derivative of A with respect to y and setting it equal to zero:

dA/dy = (-140π + 8π^2y)/16 + (2πy) = 0

Simplifying this equation, we have:
-140π + 8π^2y + 32πy = 0

Combining like terms, we get:
-140π + 40πy(2 + π) = 0

Dividing both sides by 40π, we have:
y(2 + π) = 7

Dividing both sides by (2 + π), we get:
y = 7/(2 + π)

Substituting this value of y back into equation 1, we can solve for x:
4x + 2π(7/(2 + π)) = 35

Simplifying this equation and solving for x, we find:
x = (35 - 14π/(2 + π))/4

Now that we have the values of x and y, we can calculate the circumference of the circle:

Circumference = 2πy = 2π(7/(2 + π))

Using this formula, we can calculate the circumference of the circle when A is at a minimum.

square

side = s, area = s^2
circle
circumference = c, area = pi d^2/4 = c^2/(4pi)

4s + c = 35 so c = (35-4s)
a = s^2 + c^2/(4pi)
a = s^2 + (35-4s)^2/(4pi)
a = s^2 + (1225-280s+16s^2)/12.56
a = 2.27s^2 - 22.3 s + 97.5

that is a parabola, find the vertex or use calculus (I used calculus)
0 = 4.54 s -22.3
s = 4.91
c = 35-4*4.91 = 15.3