Awire 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)

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

your answer is a

step-by-step explanation:

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