You have a wire that is 44 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?

Respuesta :

Answer:

The circumference of the circle is 29.36 cm

Step-by-step explanation:

Let

x -----> the length of first piece (shape of square)

y ------> the length of the other piece (shape of a circle)

we know that

[tex]x+y=44[/tex]

[tex]y=44-x[/tex] -----> equation A

step 1

Find out the area of square

The perimeter of square is equal to the length of the first piece

[tex]P=4b[/tex]

where

b is the length side of square

[tex]P=x\ cm[/tex]

[tex]x=4b[/tex]

[tex]b=x/4[/tex]

Find the total area A

The area of square is

[tex]A_1=b^2[/tex]

[tex]A_1=\frac{x^{2}}{16}[/tex]

step 2

Find out the area of the circle

The circumference of the circle is equal to the length of the other piece

[tex]C=(44-x)\ cm[/tex]

The circumference is equal to

[tex]C=2\pi r[/tex]

so

[tex]2\pi r=(44-x)\ cm[/tex]

Find the radius of the circle

[tex]r=\frac{(44-x)}{2\pi}\ cm[/tex]

Find the area of the circle

[tex]A_2=\pi r^{2}[/tex]

substitute the value of r

[tex]A_2=\pi (\frac{(44-x)}{2\pi})^{2}[/tex]

[tex]A_2= \frac{(44-x)^2}{4\pi}[/tex]

step 3

Find out the total area

[tex]A=A_1+A_2[/tex]

substitute

[tex]A=\frac{x^{2}}{16}+\frac{(44-x)^2}{4\pi}[/tex]

This is a vertical parabola open upward

The vertex is a minimum

using a graphing tool

The vertex is the point (14.64, 67.77)

see the attached figure

For x=14.64 cm -----> the area is a minimum

The lengths of the wire are

[tex]x=14.64\ cm\\y=44-14.64=29.36\ cm[/tex]

therefore

The circumference of the circle is 29.36 cm

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