A piece of wire 9 m long is cut into two pieces. One piece is bent into the shape of a circle of radius r and the other is bent into a square of side s. How should the wire be cut so that the total area enclosed is:

I need help finding the radius and side of the maximum and minimum. Thanks!

I will go with your definitions

then 2πr + 4s = 9
s = (9-2πr)/4
total area
= A
= πr^2 + s^2
= πr^2 + (9-2πr)^2 /16

dA/dr = 2πr + (1/8)(9-2πr)(-2π)
= 2πr - π(9-2πr)/4
= 0 for a max/min

2πr = π(9-2πr)/4
8πr = π(9-2πr)
8r = 9-2πr
8r + 2πr = 9
r(8+2π) = 9
r = 9/(8+2π) = .6301115.. m
s = 1.26022.. m

so it should be cut at the 4s mark
or at 5.04 m for the square, and 3.96 m for the circle to obtain a minimum area
The maximum of course is obtained when all the wire is used for the circle

Well, let's embark on a wire-tastic adventure, shall we?

To maximize the total area enclosed, we need to optimize the shapes of both the circle and the square.

Let's start with the circle. The length of the wire used for the circle is equal to the circumference of the circle, which is 2πr. So, we have 2πr = 9 - s. Solving for r, we get r = (9 - s)/(2π).

Now, for the square. The length of the wire used for the square is equal to four times the length of one side, which is 4s. So, we have 4s = 9 - (9 - s) = s. Solving for s, we get s = 9/5.

To find the radius, we can substitute the value of s back into our equation for r: r = (9 - s)/(2π) = (9 - 9/5)/(2π) = (36/5)/(2π) = 36/(10π) = 18/(5π).

So, the radius of the circle that maximizes the total area enclosed is 18/(5π), and the side of the square is 9/5.

As for the minimum area, if you cut the wire into two equal parts, you'll end up with a circle with a radius of (9/2π), and a square with a side length of 9/4.

Remember, wires are flexible, so they can always bend to your whims, just like my sense of humor! Good luck with your shapes!

To find the radius and side length that give the maximum and minimum total enclosed area, we need to express the area in terms of one variable and then find the maximum and minimum of that function.

Let's start by expressing the circumference of the circle, which is equal to the length of the wire (9 m), in terms of the radius (r):

C = 2πr

Now, let's express the perimeter of the square, which is also equal to the length of the wire (9 m), in terms of the side length (s):

P = 4s

To find the radius and side length that maximize the total area, we need to consider that the perimeter of both shapes must equal the length of the wire.

So, we have two equations:
2πr = 9 (equation 1)
4s = 9 (equation 2)

From equation 1, we can solve for r:
r = 9/(2π)

Substituting this value of r into equation 2, we can solve for s:
4s = 9
s = 9/4

Now we have the values of r and s that give the maximum total area.

To find the radius and side length that minimize the total area, we need to consider that the perimeter of both shapes must equal the length of the wire.

Again, we have two equations:
2πr = 9 (equation 1)
4s = 9 (equation 2)

Solving equation 1 for r:
r = 9/(2π)

Substituting this value of r into equation 2, we can solve for s:
4s = 9
s = 9/4

So, the radius and side length that give both the maximum and minimum total enclosed area are:

Radius (r): 9/(2π)
Side length (s): 9/4

However, it's worth noting that when the wire is divided into two pieces with these lengths, the total areas of the circle and the square will be equal.

To find the radius and side length that give the maximum and minimum total enclosed area, we can use calculus.

Let's denote the radius of the circle as r and the side length of the square as s.

The circumference of the circle is given by 2πr, and the perimeter of the square is 4s. Since the total length of the wire is 9 m, we can write the equation:

2πr + 4s = 9

We want to find the maximum and minimum areas, so we need to express the area in terms of either r or s.

The area of the circle is given by A_circle = πr^2, and the area of the square is A_square = s^2. So the total enclosed area is:

A_total = πr^2 + s^2

To eliminate one variable and express the area only in terms of the other variable, we can solve the first equation for s and substitute it into the area equation:

s = (9 - 2πr) / 4

Substituting this expression for s into the area equation, we get:

A_total = πr^2 + ((9 - 2πr) / 4)^2

Now, with A_total expressed only in terms of r, we can find the derivative dA_total/dr and set it equal to zero to find critical points where the area is at its maximum or minimum.

Taking the derivative and simplifying, we get:

dA_total/dr = 2πr - (9 - 2πr)(2π/4) = 2πr - π(9 - 2πr)/2 = πr - 9π/2 + π^2r/2

Setting dA_total/dr equal to zero, we have:

πr - 9π/2 + π^2r/2 = 0

Simplifying further, we get:

r(π + π^2/2) = 9π/2

Dividing both sides by (π + π^2/2), we find:

r = (9π/2) / (π + π^2/2)

Simplifying, we get:

r = (9/2) / (1 + π/2)

Now that we have the value of r, we can substitute it back into the equation for s to find its value.

s = (9 - 2πr) / 4

s = (9 - 2π[(9/2) / (1 + π/2)]) / 4

Calculating these values, we obtain the corresponding radius and side length.