You have a wire that is 77 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?

If the square has side s and the circle has radius r, then

A = s^2 + πr^2
But, 4s+2πr = 77, so s = (77-2πr)/4
A(r) = (77-2πr)^2/16 + πr^2 = (π62/4 + π)r^2 - 77π/4 r + 5929/16
The vertex of that parabola is at r = 77/(8+2π)
C = 2πr = 77π/(4+π)

To find the circumference of the circle when A is a minimum, we need to determine the values of the lengths of the square and the circle.

Let's represent the length of the square as x. Since all four sides of a square are equal, the perimeter of the square is equal to 4x. Hence, one side of the square will be x/4.

Now, let's represent the length of the wire used to form the circle as y. The circumference of the circle is given by the formula 2πr, where r is the radius of the circle. Since the whole wire of length y is used to form the circle, the circumference is equal to y.

We have the equation: x/4 + y = 77

To find the minimum A, we need to determine the values of x and y that satisfy this equation and minimize the area.

To proceed, we need to eliminate one variable from the equation. Let's solve the equation for y:

y = 77 - x/4

Next, we substitute the value of y in the circumference formula:

C = 2πr = 2π(77 - x/4)

Finally, to find the value of x that minimizes the area, we differentiate A with respect to x and set it to zero:

dA/dx = d[πr^2 + (x/4)^2]/dx = 2πr(dr/dx) + 2(x/4)(1/4) = 2πr(dr/dx) + x/8

dA/dx = 0 implies that 2πr(dr/dx) + x/8 = 0

Substituting the equation for r = (y/2π), we have:

2π(y/2π)(dr/dx) + x/8 = 0
y(dr/dx) + x/8 = 0
(dr/dx) = -x/8y

Now, substituting the equation for y = 77 - x/4:

c = -x/8(77 - x/4)

Simplifying, we have:

c = -77x/8 + x^2/32

To find the value of x that minimizes the circumference, we differentiate c with respect to x and set it to zero:

dc/dx = -77/8 + x/16 = 0

Rearranging the equation, we have:

x/16 = 77/8
x = (77/8) * 16
x = 77 * 2 = 154

Substituting x = 154 into the equation for y = 77 - x/4:

y = 77 - 154/4
y = 77 - 38.5 = 38.5

Therefore, the lengths of the square and the circle are x = 154 cm and y = 38.5 cm, respectively.

Finally, substituting these values into the formula for the circumference of the circle:

C = 2πr = 2π(38.5/2) = 2π(19.25) = 38.5π cm

Thus, the circumference of the circle when A is a minimum is 38.5π cm.

To find the circumference of the circle when the total area (A) is a minimum, we need to first express the area (A) in terms of a single variable. Let's represent the length of one side of the square as x and the radius of the circle as r.

Since the wire is cut into two pieces, one for the square and one for the circle, and the total length of the wire is 77 cm, we have: 4x + 2πr = 77.

The area of the square is given by A(square) = x^2, and the area of the circle is given by A(circle) = πr^2. So, the total area is A = A(square) + A(circle) = x^2 + πr^2.

To find the minimum value of A, we can take the derivative of A with respect to x (or r), set it equal to zero, and solve for x (or r), as the derivative will give us the critical points.

Taking the derivative of A with respect to x:
dA/dx = 2x + 0 (since the derivative of πr^2 with respect to x is zero)
Setting dA/dx = 0:
2x = 0
x = 0

Similarly, taking the derivative of A with respect to r:
dA/dr = 0 + 2πr
Setting dA/dr = 0:
2πr = 0
r = 0

However, since a side length or radius cannot be zero for a square or circle respectively, these values are not valid critical points.

To find the minimum value of A, we must consider the endpoints of the possible range of values for x and r. In this case, since x represents a side length of the square, it must be greater than zero. For r, the radius of the circle, it must also be greater than zero.

Therefore, to minimize A, we need to find the values of x and r that satisfy the constraint 4x + 2πr = 77, while also ensuring that x > 0 and r > 0.

Solving the equation 4x + 2πr = 77 for one variable will allow us to then substitute that solution into the area equation A = x^2 + πr^2 and solve for the other variable.

Let's solve for x in terms of r:
4x = 77 - 2πr
x = (77 - 2πr)/4

Substituting this value of x into A, we get:
A = [(77 - 2πr)/4]^2 + πr^2

Next, differentiate A with respect to r:
dA/dr = 2(77 - 2πr)/4 * (-2π/4) + 2πr
simplifying,
dA/dr = (-2π(77 - 2πr) + 4π^2r)/8 + 2πr
dA/dr = (-154π + 4π^2r + 16πr)/8 + 2πr
dA/dr = (-154π + 4π^2r + 16πr + 16πr)/8
dA/dr = (-154π + 32πr)/8
dA/dr = (-19π + 4rπ)

Setting dA/dr = 0:
-19π + 4rπ = 0
4rπ = 19π
r = 19/4

Since r > 0, we know that r = 19/4 is a valid critical point.

Substituting r = 19/4 into the constraint equation 4x + 2πr = 77 to find x:
4x + 2π(19/4) = 77
4x + 19π/2 = 77
4x = 77 - 19π/2
x = (77 - 19π/2)/4

Now that we have the values of x and r that satisfy the constraint and ensure a minimum value for A, we can find the circumference of the circle.

Since the circumference of a circle is given by C = 2πr, we can substitute the value of r = 19/4 into the formula:
C = 2π(19/4)
C = 19π/2

Therefore, the circumference of the circle, when the total area (A) is a minimum, is 19π/2.