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)

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