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A piece of wire 24 ft long is cut into two pieces. One piece is used to form a square, and the remaining piece is used to form a circle. Where should the wire be cut so that the combined area of the two figures is a maximum? (Round your answers to two decimal places.)

length of the wire used for the square = ? ft

length of the wire used for the circle = ? ft

Where should the wire be cut so that the combined area of the two figures is a minimum? (Round your answers to two decimal places.)

length of the wire used for square = ? ft

length of the wire used for circle = ? ft

Respuesta :

Answer:

SAY WHAT

Step-by-step explanation:

MrRoyal

The area of a shape is the amount of space it occupies.

  • The length of wire used for the square is 13.45 ft
  • The length of wire used for the circle is 10.55 ft

The given parameter is:

[tex]\mathbf{l = 24}[/tex] -- the length of the original wire

Assume the length that makes up the square is x.

So, we have:

[tex]\mathbf{y = \frac x4}[/tex] --- length of one side

[tex]\mathbf{A_1 = \frac{x^2}{16}}[/tex] --- area of the square

The circumference of the circle would be:

[tex]\mathbf{C = 24 - x}[/tex]

The circumference is calculated using:

[tex]\mathbf{C = 2\pi r}[/tex]

So, we have:

[tex]\mathbf{2\pi r = 24 - x}[/tex]

Make r the subject

[tex]\mathbf{r = \frac{24 - x}{2\pi }}[/tex]

The area of the circle is:

[tex]\mathbf{A_2 = \pi \times (\frac{24 - x}{2\pi })^2}[/tex]

[tex]\mathbf{A_2 = \frac{(24 - x)^2}{4\pi }}[/tex]

So, the area of the complete wire is:

[tex]\mathbf{A = A_1 + A_2}[/tex]

[tex]\mathbf{A = \frac{x^2}{16} + \frac{(24 - x)^2}{4\pi }}[/tex]

Expand

[tex]\mathbf{A = \frac{x^2}{16} + \frac{576 -48x + x^2}{4\pi }}[/tex]

Differentiate

[tex]\mathbf{A' = \frac{2x}{16} + \frac{-48 + 2x}{4\pi }}[/tex]

[tex]\mathbf{A' = \frac{x}{8} + \frac{-24 + x}{2\pi }}[/tex]

Set to 0

[tex]\mathbf{\frac{x}{8} + \frac{-24 + x}{2\pi } = 0}[/tex]

Split

[tex]\mathbf{\frac{x}{8} - \frac{24}{2\pi} + \frac{x}{2\pi } = 0}[/tex]

Collect like terms

[tex]\mathbf{\frac{x}{8} + \frac{x}{2\pi } = \frac{24}{2\pi}}[/tex]

[tex]\mathbf{\frac{x}{8} + \frac{x}{2\pi } = \frac{12}{\pi}}[/tex]

Take LCM

[tex]\mathbf{\frac{x\pi + 4x}{8\pi} = \frac{12}{\pi}}[/tex]

Multiply both sides by [tex]8\pi[/tex]

[tex]\mathbf{x\pi + 4x = 96}[/tex]

Factor out x

[tex]\mathbf{x(\pi + 4) = 96}[/tex]

Solve for x

[tex]\mathbf{x = \frac{96}{\pi + 4}}[/tex]

[tex]\mathbf{x = \frac{96}{3.14 + 4}}[/tex]

[tex]\mathbf{x = \frac{96}{7.14}}[/tex]

[tex]\mathbf{x = 13.45}[/tex]

Recall that:

[tex]\mathbf{C = 24 - x}[/tex]

[tex]\mathbf{C = 24 - 13.45}[/tex]

[tex]\mathbf{C = 10.55}[/tex]

This means that:

  • The length of wire used for the square is 13.45 ft
  • The length of wire used for the circle is 10.55 ft

Read more about minimizing areas at:

https://brainly.com/question/13784786

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