A 100 inch piece of wire is divided into 2 pieces and each piece is bent into a square. How should this be done in order of minimize the sum of the areas of the 2 squares?

a) express the sum of the areas of the squares in terms of the lengths of x and y of the 2 pieces
b) what is the constraint equation relating x and y?
c) does this problem require optimization over anopen or closed interval?
d) solve the optimization problem

Let one piece to shape the first square be 4x, then let the other piece be 4y

4x + 4y = 100
x+y = 25
y = 25-x

Sum of areas = x^2 + y^2
= x^2 + (25-x)^2
= 2x^2 - 50x + 625
d(Sum of areas)/dx = 4x - 50
= 0 for a max/min of the sum of the areas
4x = 50
x = 12.5

( I defined the length as 4x instead of x to avoid fractions)

a) To express the sum of the areas of the squares in terms of the lengths of x and y, we need to find the formula for the areas of each square.

Let's assume that x is the length of the first piece of wire, and y is the length of the second piece of wire.

The perimeter of a square is given by 4 times the length of one side. Since each piece of wire is bent into a square, the length of each side of the squares will be equal to the lengths of their respective pieces.

So, the formulas for the areas of the squares will be:

Area of the first square = x^2
Area of the second square = y^2

Therefore, the sum of the areas of the squares is:

Sum of areas = x^2 + y^2

b) The constraint equation relating x and y is that the total length of the wire used should be equal to 100 inches. Mathematically, this can be expressed as:

x + y = 100

c) This problem requires optimization over a closed interval. Since we are dealing with lengths, x and y must be non-negative and finite. This means that the possible values for x and y are within the closed interval [0, 100].

d) To solve the optimization problem, we need to minimize the sum of the areas of the squares subject to the constraint equation x + y = 100.

To do this, we can substitute y = 100 - x into the equation for the sum of the areas:

Sum of areas = x^2 + (100 - x)^2

Expanding and simplifying the expression, we get:

Sum of areas = x^2 + 10000 - 200x + x^2

Combining like terms, we have:

Sum of areas = 2x^2 - 200x + 10000

To find the minimum of this quadratic function, we can take the derivative and set it equal to zero:

d/dx (Sum of areas) = 4x - 200 = 0

Solving for x, we get:

4x = 200
x = 50

To find y, we can substitute the value of x into the constraint equation:

x + y = 100
50 + y = 100
y = 50

So, the lengths of the two pieces of wire that minimize the sum of the areas of the squares are x = 50 inches and y = 50 inches.

a) Let's express the sum of the areas of the squares in terms of the lengths of x and y of the two pieces.

The length of one piece is x, and the length of the other piece is y. Each piece is bent into a square.

The perimeter of a square is equal to 4 times its side length. Since each piece forms a square, the perimeter of the first square is 4x, and the perimeter of the second square is 4y.

The sum of the perimeters of the two squares should equal the total length of the wire, which is 100 inches. So we can write the constraint equation as:

4x + 4y = 100

To find the sum of the areas, we need to find the side lengths of each square. The side length of the first square can be found by dividing the perimeter by 4: x = (4x)/4 = x. Similarly, the side length of the second square is y.

The area of a square is equal to the square of its side length. So the area of the first square is x^2, and the area of the second square is y^2.

The sum of the areas of the two squares is:

x^2 + y^2

b) The constraint equation relating x and y is 4x + 4y = 100, as mentioned above.

c) This problem requires optimization over a closed interval. Since x and y represent lengths, they cannot be negative or undefined. They must be non-negative. Additionally, the sum of the lengths of the two pieces of wire should be equal to the total length of the wire, which is fixed at 100 inches.

d) To solve the optimization problem, we need to minimize the sum of the areas, which is x^2 + y^2, subject to the constraint equation 4x + 4y = 100.

First, let's rearrange the constraint equation to solve for y:

4y = 100 - 4x
y = (100 - 4x) / 4
y = 25 - x

Substitute this expression for y into the equation for the sum of the areas:

sum_of_areas = x^2 + (25 - x)^2

To minimize this function, we can take the derivative with respect to x and set it equal to zero:

d(sum_of_areas) / dx = 2x - 2(25 - x) = 2x - 50 + 2x
4x - 50 = 0

Solve this equation for x:

4x = 50
x = 12.5

Now substitute this value of x back into the constraint equation to find y:

4(12.5) + 4y = 100
50 + 4y = 100
4y = 50
y = 12.5

So the optimal lengths for the two pieces of wire are x = 12.5 inches and y = 12.5 inches, resulting in two squares with side lengths of 12.5 inches.

The minimum sum of the areas of the squares is then:

sum_of_areas = (12.5)^2 + (12.5)^2 = 156.25 + 156.25 = 312.5 square inches