A piece of wire 28 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.
(a) How much wire should be used for the square in order to maximize the total area? (b) How much wire should be used for the square in order to minimize the total area?
14/(pi +4)
To solve this problem, we need to use the concepts of optimization and calculus. Let's break down the problem step-by-step:
Step 1: Define the variables
Let's assume that x meters of wire are used to create the square, and the remaining wire (28 - x) meters are used to create the circle.
Step 2: Calculate the side length of the square
Since the wire is bent into a square, the perimeter of the square is equal to the wire length:
Perimeter of square = 4 * side length = x
Therefore, the side length of the square is:
Side length of square = x / 4
Step 3: Calculate the radius of the circle
Since the remaining wire is bent into a circle, the circumference of the circle is equal to the remaining wire length:
Circumference of circle = 2 * π * radius = 28 - x
Therefore, the radius of the circle is:
Radius of circle = (28 - x) / (2 * π)
Step 4: Calculate the area of the square
The area of a square is given by:
Area of square = Side length * Side length
Substituting the value of the side length of the square obtained in Step 2:
Area of square = (x / 4)^2 = x^2 / 16
Step 5: Calculate the area of the circle
The area of a circle is given by:
Area of circle = π * radius * radius
Substituting the value of the radius of the circle obtained in Step 3:
Area of circle = π * ((28 - x) / (2 * π))^2 = (28 - x)^2 / (4 * π)
Step 6: Find the total area function
The total area is the sum of the area of the square and the area of the circle:
Total area = Area of square + Area of circle = x^2 / 16 + (28 - x)^2 / (4 * π)
Step 7: Find the derivative of the total area function
To find the maximum or minimum of the total area, we need to find the critical points of the function. Taking the derivative of the total area function from Step 6 with respect to x yields:
d(Total area)/dx = 2x / 16 - 2(28 - x) / (4 * π)
Step 8: Set the derivative equal to zero and solve for x
Setting the derivative obtained in Step 7 equal to zero gives:
2x / 16 - 2(28 - x) / (4 * π) = 0
Simplifying:
x / 8 - (28 - x) / (2 * π) = 0
Multiplying through by 8 * π:
πx - 4π(28 - x) = 0
πx - 112π + 4πx = 0
5πx - 112π = 0
x = 112π / 5
Step 9: Determine the maximum and minimum areas
Substituting the value of x obtained in Step 8 back into the total area function from Step 6 gives the maximum and minimum areas.
(a) Maximum area:
Substituting x = 112π / 5 into the total area function:
Total area = (112π / 5)^2 / 16 + (28 - 112π / 5)^2 / (4 * π)
Simplifying:
Total area ≈ 1160.52 square meters
(b) Minimum area:
The minimum area occurs at the end points of the feasible interval, which are x = 0 and x = 28. Substituting these values into the total area function:
At x = 0: Total area = 0^2 / 16 + (28 - 0)^2 / (4 * π) ≈ 64.36 square meters
At x = 28: Total area = 28^2 / 16 + (28 - 28)^2 / (4 * π) ≈ 196 square meters
Therefore, to (a) maximize the total area, 112π / 5 meters of wire should be used for the square, and to (b) minimize the total area, x = 0 or 28 meters of wire should be used for the square.
To solve this problem, we can use the concept of optimization, where we aim to maximize or minimize a certain quantity given a specific condition. In this case, we want to find the amount of wire to be used for the square to maximize or minimize the total area.
Let's break down the problem step by step:
(a) Maximizing the total area:
To maximize the total area, we need to find the value of the side length of the square that uses up the most wire. Let's assume that the length of the wire used for the square is denoted as "x." We know that the total wire length is 28 m, so the length of wire used for the circle, denoted as "y," would be 28 - x.
Now, let's find the relationship between the side length of the square and its perimeter (the amount of wire used for the square). Since a square has all sides equal in length, the perimeter of the square is 4 times the length of one side. Therefore, we have:
Perimeter of the square = 4x
Given that the total wire length is 28 m, we have:
4x + y = 28
To find the total area, we need to consider both the square and the circle. The area of a square is equal to the square of the length of one side, and the area of a circle is given by the equation πr^2, where r is the radius. Since the wire for the circle is the remaining length (28 - x), the radius of the circle would be (28 - x) / (2π).
Therefore, the total area can be expressed as:
Total Area = x^2 + π((28 - x) / (2π))^2
To find the maximum total area, we differentiate the Total Area equation with respect to x and set it equal to zero. Then, solve for x:
Total Area' = d(Total Area)/dx = 2x - 2(28 - x) / (2π)
Setting it equal to zero:
2x - 2(28 - x) / (2π) = 0
Simplifying the equation:
2x = (28 - x) / π
Multiplying both sides by π:
2πx = 28 - x
Rearranging terms:
3x = 28
Solving for x:
x = 28 / 3
Now we have the value of x, which represents the amount of wire used for the square that maximizes the total area. To find the exact value, we substitute it back into the Total Area equation:
Total Area = (28 / 3)^2 + π((28 - (28 / 3)) / (2π))^2
= 784 / 9 + π(56 / (3 * 2π))^2
= 784 / 9 + (28 / 3)^2
= 784 / 9 + 784 / 9
= 1568 / 9
Therefore, the wire length for the square that maximizes the total area is 28/3 m, and the maximum total area is 1568/9.
(b) Minimizing the total area:
To minimize the total area, we follow a similar approach as above, but this time, we minimize the total area by finding the value of x that gives the smallest total area.
To begin, we rewrite the equation relating the wire lengths for the square and the circle:
4x + y = 28
Next, we use the same formula for the total area:
Total Area = x^2 + π((28 - x) / (2π))^2
Differentiating and setting the equation equal to zero, we find:
2x - 2(28 - x) / (2π) = 0
After simplifying, we get:
2x = (28 - x) / π
Multiplying both sides by π:
2πx = 28 - x
Rearranging terms:
3x = 28
Solving for x:
x = 28 / 3
Substituting it back into the Total Area equation:
Total Area = (28 / 3)^2 + π((28 - (28 / 3)) / (2π))^2
= 784 / 9 + π(56 / (3 * 2π))^2
= 784 / 9 + (28 / 3)^2
= 784 / 9 + 784 / 9
= 1568 / 9
Therefore, the wire length for the square that minimizes the total area is also 28/3 m, and the minimum total area is 1568/9.
a) that part is trivial, the largest area is obtained if all of the wire is used to make a circle
b) let the radius of the circle be r
let the side of the square be x
4x + 2πr = 28
2x + πr = 14
x = (14-πr)/2
A = x^2 + πr^2
= (14-πr)^2 /4 + πr^2
dA/dr = (1/2)(14-πr)(-π) + 2πr = 0 for a max/min
simplifying ...
4r = 14 - πr
4r + πr = 14
r = 14/(4+π)
we need 2πr to form the circle, so the wire should be cut at
2π(14/(4+π)) or appr 12.32 m
check my arithmetic