A wire 9 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each:

For the square?
For the circle?

Where should the wire be cut to maximize the total area? Again, give the length of wire used for each:
For the square?
For the circle?

To figure out where the wire should be cut to minimize the total area, we can use the concept of optimization. Let's start by assigning variables to the length of each piece.

Let 'x' represent the length of the wire used for the square, and '9 - x' represent the length used for the circle.

For the square:
The square requires 4 equal sides, so each side will have a length of (x/4). This means the perimeter of the square is 4 * (x/4) = x.

For the circle:
The circumference of a circle is given by the formula C = 2πr, where 'C' represents the circumference, and 'r' is the radius. In this case, the circumference is given by 9 - x.

To minimize the total area, we need to minimize the sum of the areas of the square and the circle. The area of the square is A = side^2, and the area of the circle is A = π * r^2.

Now, let's calculate the area of the square and the circle:

Area of the square = (x/4)^2 = x^2/16
Area of the circle = π * (r^2) = π * ((9 - x) / (2π))^2 = (9 - x)^2 / (4π)

The total area is the sum of the areas of the square and the circle:

Total Area = (x^2/16) + (9 - x)^2 / (4π)

To find the minimum total area, we can take the derivative of the Total Area equation with respect to 'x' and set it equal to zero:

d(Total Area) / dx = (1/8) * x - ((9 - x)^2) / (8π) = 0

To solve this equation, we can simplify it first:

x/(8/π) - (81 - 18x + x^2) / (8π) = 0
x/(8/π) = (81 - 18x + x^2) / (8π)
πx = 81 - 18x + x^2

Rewriting the equation in quadratic form:
x^2 - 18x + 81 - πx = 0

To find the minimum or maximum value, we can now solve this quadratic equation.

Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), where a = 1, b = -18, and c = 81 - π.

Calculating the discriminant, b^2 - 4ac:
(-18)^2 - 4 * 1 * (81 - π) = 324 - 4(81 - π) = 324 - 324 + 4π = 4π

Since the discriminant 4π is positive, there are two real solutions for x, one for minimum and the other for maximum.

Solving for x:
x = (-(-18) ± √(4π))/2 = (18 ± 2√π)/2 = 9 ± √π

Now, we need to determine if these values are within the range of 0 to 9. Since π > 3, 9 + √π > 9 + √3 > 9 + 1.7 ≈ 10.7, which is outside the range. Therefore, we can discard this value.

The other value, 9 - √π, falls within the range. So, we conclude that the wire should be cut at a length of 9 - √π meters for the square and √π meters for the circle to minimize the total area.

To find the maximum total area, we can use the same logic, and note that the maximum total area will occur when the wire is cut at the remaining length (9 - (9 - √π)) meters for the square and (9 - √π) meters for the circle.

Therefore, for the maximum total area:
For the square: 9 - (9 - √π) = √π meters
For the circle: 9 - √π meters

n-7w

A wire 60 inches long is to be cut into two pieces. One of the pieces will be bent into the shape of the circle and the other into the shape of an equilateral triangle. What is the maximum sum of the areas in sq. inch of the circle and triangle possible?

total area= PIr^2+ ((9-2r)/4)^2

so find r when dA/dr=0

lengths of two pieces: 2r, 9-2r

you will get two solutions, one for min area, one for max area.