Mathematics, 08.11.2019 05:31 kmc3490

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.)

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Answer from: Ideen

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 = $\left(\dfrac{X}{4} \right)^2$

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

The radius of the circle is therefore;

$r = \dfrac{28 - X}{2 \cdot r}$

The area of a circle, A₂ = π·r²

Therefore;

$A_2 = \pi \cdot \left(\dfrac{28 - X}{2 \cdot \pi} \right)^2 = \dfrac{\left(28 - X\right)^2}{4 \cdot \pi}$

The area of the circle, A₂ = $\dfrac{(28 - X)^2}{4 \cdot \pi}$

The sum of the area is therefore;

$A = \left(\dfrac{X}{4} \right)^2 + \dfrac{(28 - X)^2}{4 \cdot \pi}$

$A = \dfrac{(\pi + 4)\cdot X^2 + 224 \cdot X+ 3136}{16 \cdot \pi}$

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;

$x = -\dfrac{b}{2 \cdot a}$

Therefore;

$X = - \dfrac{\dfrac{224}{16 \cdot \pi} }{2 \times \dfrac{\pi + 4}{16 \cdot \pi} } = \dfrac{112}{\pi + 4} \approx 15.68$

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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Answer from: christellechery11

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 = $\frac{l}{4}$

So, the area of the square = $\frac{l^{2}}{16}$ ft²

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

2πr = (28 - l)

r = $\frac{28-l}{2\pi }$

Area of the circle formed = πr²

= $\frac{\pi(28-l)^{2} }{4(\pi )^{2} }$

= $\frac{(28-l)^{2}}{4\pi }$

Combined area of the square and circle

= $\frac{l^{2}}{16}$ + $\frac{(28-l)^{2}}{4\pi }$

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

$\frac{dA}{dl}=\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28-l)^{2}}{4\pi}]$

= $\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28)^{2}+l^{2}-56l }{4\pi }]$

= $\frac{2l}{16}+\frac{(2l-56)}{4\pi }$

Now equate the derivative to zero.

$\frac{2l}{16}+\frac{(2l-56)}{4\pi }$ = 0

$(2l-56)=-\frac{2l}{16}\times 4\pi$

$(2l-56)=-\frac{\pi l}{2}$

$2l+\frac{\pi l}{2}=56$

$\frac{(4l+l\pi )}{2}=56$

l(4 + π) = 112

l(4 + 3.14) = 112

l = $\frac{112}{7.14}$

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

Apiece of wire 28 ft long is cut into two pieces. one piece is used to form a square, and the remain

Answer from: Quest

bull

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.

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