A wire 17 inches long is cut into two pieces. One piece is bent to form a square and the other is bent to form a rectangle that is twice as wide as it is high. How should the wire be cut in order to minimize the total area of the square and rectangle?

4 s + 2w + 4w = 4 s + 6 w = 17

A = s^2 + w(2w) = s^2 + 2 w^2
s = (17-6w)/4

A = (1/16)(17-6w)^2 + 2 w^2

dA/dw = 0 at min

0 = (1/8)(17-6w)(-6) = 4 w

(3/4)(17-6w) = 4 w

3 (17-6w) = 16 w

51 - 18 w = 16 w

34 w = 51

w = 51/34 = 1.5
then find s and w and 2 w

Let's denote the length of the wire used to form the square as x.

Since the perimeter of a square is given by 4 times the length of one side, we have:
4x = x

Simplifying the equation, we find:
4x = 17
x = 17/4
x ≈ 4.25 inches

Therefore, we should cut the wire into two pieces: one with a length of approximately 4.25 inches, and the other with a length of 17 - 4.25 = 12.75 inches.

Now let's calculate the dimensions of the rectangle using the second piece of wire.

Let the height of the rectangle be h, and its width be 2h (twice as wide as it is high).

Since the perimeter of a rectangle is given by twice the sum of the length and width, we have:
2h + 2(2h) = 12.75

Simplifying the equation, we find:
6h = 12.75
h ≈ 2.125 inches

So, the height of the rectangle is approximately 2.125 inches, and the width is 2h ≈ 4.25 inches.

To minimize the total area of the square and rectangle, the wire should be cut into pieces measuring approximately 4.25 inches and 12.75 inches, which will result in a square with a side length of approximately 4.25 inches and a rectangle with dimensions of approximately 2.125 inches by 4.25 inches.

To minimize the total area of the square and rectangle, we need to find the values that will minimize their individual areas.

Let's begin by finding the formulas for the area of the square and rectangle.

1. Area of a square: A = s^2, where s is the side length of the square.
2. Area of a rectangle: A = w * h, where w is the width and h is the height of the rectangle.

Now, let's determine the dimensions of the square and rectangle in terms of the total length of the wire, which is 17 inches.

Let's assume that the length of the wire used to form the square is x. Since a square has four equal sides, each side will have a length of x/4.

The remaining length of the wire not used for the square will be (17 - x).

According to the problem, the rectangle is twice as wide as it is high. Therefore, the width of the rectangle can be written as (2h), and the height can be written as h.

Now, we have the following relationships:

Length of the wire used for the square (x) = 4 * (side length of the square) = 4 * (x/4) = x.
Length of the wire used for the rectangle = 17 - x.

Next, we can deduce the expressions for the side length of the square (s) and the dimensions of the rectangle (w and h):

s = x/4
w = 2h
h = (17 - x)/2

To find the total area (T) of the square and rectangle, we can substitute these expressions into our area formulas:

T = A(square) + A(rectangle)
T = s^2 + w * h
T = (x/4)^2 + (2h)((17 - x)/2)

Simplifying the equation:

T = x^2/16 + h(17 - x)

Now, to minimize the total area T, we need to find the value of x that minimizes the equation.

To do this, we can take the derivative of T with respect to x, set it equal to zero, and solve for x:

d(T)/d(x) = 2x/16 - h = 0
2x/16 = h

Now, substitute the value of h in terms of x:

2x/16 = (17 - x)/2

Solving for x:

2x = (17 - x)(16)
2x = 272 - 16x
18x = 272
x = 272/18
x ≈ 15.11

Since we are dealing with lengths, we need to round x to the nearest whole number, giving us x = 15.

So, to minimize the total area of the square and rectangle, the wire should be cut into two pieces: one with a length of 15 inches for the square, and the other with a length of (17 - 15) = 2 inches for the rectangle.

There are several similar related questions below. I'm sure you can adapt one of those solutions to your specific numbers.