Apiece of wire 28 ft long is cut into two pieces. one piece is used to form a square, and the remaining piece is used to form a circle. where should the wire be cut so that the combined area of the two figures is a maximum? (round your answers to two decimal places.)

The combined area formed by the circle and square will give U-shaped

parabolic curve, and therefore, will only have a minimum point, not a

maximum point in between

So that the combined area will be a minimum, the wire should be cut into

15.68 ft. and 12.32 ft. length pieces

Reason:

Let X represent the length of the piece used for the square;

The length of the piece used for the circle is therefore 28 - X

The area of the square, A₁ =

Circumference of the circle, C = 28 - X = 2·π·r

The radius of the circle is therefore;

The area of a circle, A₂ = π·r²

Therefore;

The area of the circle, A₂ =

The sum of the area is therefore;

From the above quadratic equation of the combined the coefficient of X²,

therefore, the graph has only a minimum point if the wire is used for both a

square and a circle.

At the minimum point, we have;

Therefore;

The length of the wire to be used as a square to get the minimum area, X ≈

15.68 ft.

The length used for the circle is therefore; 28 ft. - 15.68 ft. ≈ 12.32 ft.

Therefore, the wire should be cut into 15.68 ft. and 12.32 ft. pieces.

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Wire should be cut in two parts with the length = 15.69 ft and 12.31 ft

Step-by-step explanation:

Length of the wire = 28 ft has been cut in two pieces.

One piece is used to form a square and remaining piece to form a circle.

Let the length of the wire which forms the square is 'l' ft.

Area of the square = (side)²

Perimeter of the square = 4(side) = l

Length of one side =

So, the area of the square = ft²

Now length of the remaining part = perimeter of the circle = (28 - l) ft

2πr = (28 - l)

r =

Area of the circle formed = πr²

=

=

Combined area of the square and circle

= +

= +

Now to maximize the area we will find the derivative of the area with respect to l.

=

=

Now equate the derivative to zero.

= 0

l(4 + π) = 112

l(4 + 3.14) = 112

l =

l = 15.69 ft

Length of the other part = 28 - 15.69 = 12.31 ft

Therefore, wire should be cut in two parts with the length = 15.69 ft and 12.31 ft