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

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!