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A wire 6 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures?

Give the length of wire used for each:
For the square:
For the circle:
(for both, include units)

Where should the wire be cut to maximize the total area? Again, give the length of wire used for each:For the square:For the circle:(for both, include units)

Respuesta :

Answer:

Used wire in circle  x = 2.64 m

Used in square   L - x = 3.36 m

Total wire used 6 m

Step-by-step explanation:

We have a wire of 6 meters long.

We will cut it a distance x from one end, to get two pieces

x    and   6 - x

We are going to use the piece x  to get the circle then

So Perimetr of a circle is 2π*r    (r is the radius of the circle) then:

x = 2*π*r    ⇒    r = x/2*π

And area would be  A(c) = π* (x/2*π)²   ⇒ A(c) = x²/4π

From 6 - x we will get a square, and as the perimeter is 4 times the side

we have

( 6 - x )/ 4  is the side of the square

And the area is  A(s) = [( 6 - x ) /4]²

Total area as function of x is

A(t)  = A(c) + A(s)

A(x)  =  x²/4π  + [ ( 6  -  x  ) / 4 ]²

A(x)  =   x²/4π  + (36 + x² - 12x) /16

A(x)  = 1 / 16π [ 4x² + 36π + πx² -  12π x ]

Taking drivatives on both sides of the equation we get:

A´(x) = 1/ 16π [8x +2πx - 12π]

A´(x) = 0    ⇒      1/ 16π [8x +2πx - 12π]  = 0

[8x +2πx - 12π]  = 0

8x + 6.28x -  37.68  = 0

14.28x - 37.68 =  0      ⇒  x  = 37.68 /14.28

x = 2.64 m   length of wire used in the circle

Then the length L  for the side of the square is  

(6 - x )/4    ⇒ ( 6 - 2.64 )/ 4   ⇒ 3.36 / 4    

L = 0.84 m   total length of wire used in the square is

3.36 m

And total length of wire used is 6 m

The function is a quadratic  function and "a" coefficient is positive then is open upward parabola there is not a maximun

aksnkj

Answer:

Wire used in circle , x = 2.64 m

Wire used in square, L - x = 3.36 m

Total used wire is 6 m

Step-by-step explanation:

We have a wire of 6 metres long.

We will cut it a distance x metre from one end, to get two pieces x metre and 6 - x metres.

We are going to use the piece of x metre to get the circle

So, Perimeter of the circle is [tex]2\pi r[/tex] (r is the radius of the circle) then

[tex]x = 2\pi r[/tex] ⇒ [tex]r = \frac{x}{2} \pi[/tex]

And area would be [tex]A(c) =\pi (\frac{x}{2} \pi )^{2}[/tex]⇒[tex]A(c) = \frac{x^{2} }{4\pi }[/tex]

From [tex]6 - x[/tex] we will get the square, and as the perimeter is 4 times the side

we have

[tex]\frac{6 - x}{4}[/tex] is the side of the square

and the area is [tex]A(s) = (\frac{6 - x}{4}) ^{2}[/tex]

Total area of the function of x is

[tex]A(t) = A(c) + A(s)[/tex]

[tex]A(x) = \frac{x^{2} }{4\pi } +(\frac{6 - x}{4} )^{2}\\A(x) = \frac{x^{2} }{4\pi } + \frac{36+x^{2} -12x}{16} \\A(x) = \frac{1}{16\pi } (4x^{2} +36\pi +\pi x^{2} -12\pi x)[/tex]

Taking derivative on the both side of the equation we get :

[tex]A^{'} = \frac{1}{16\pi } (8x+2\pi x-12\pi )\\[/tex]

[tex]A^{'} = 0[/tex]

[tex]\frac{1}{16\pi } (8x+2\pi x-12\pi ) = 0\\(8x+2\pi x-12\pi ) = 0\\8x + 6.28x-37.68=0\\14.28x-37.68=0\\x=2.64 m[/tex]

length of wire used in the circle is x = 2.64 m

Then the length L of the wire used in the square is

[tex]\frac{6 - x}{4}[/tex] ⇒[tex]\frac{6 - 2.64}{4}[/tex] ⇒ [tex]\frac{3.36}{4}[/tex]

L = 0.84 m

Total length of the wire used in the square is 4L = 3.36 m

And total length of the wire used is 6 m

The function is a quadratic function and "a" coefficient is positive then is open upward parabola there is not a maximum.

For more information :

https://brainly.com/question/4351903?referrer=searchResults

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