Question

Wire of length 20m is divided into two pieces and the pieces are bent into a square and a circle. How should this be done in order to minimize the sum of their areas? Round your answer to the nearest hundredth?

Answer 1

Answer:

The wire must be divided into two pieces of 11.20 meters and 8.20 meters.

Step-by-step explanation:

Given,

The length of the wire = 20 meters,

Let the first part of the wire = x meters,

So, the second part of the wire = (20-x) meters,

Since, by the first part of the wire, a square is formed,

if a be the side of the square,

Then 4a = x       ( perimeter of a square = 4 × side ),

$\implies a =\frac{x}{4}$

So, the area of first part,

$A_1=(\frac{x}{4})^2=\frac{x^2}{16}$    ( area of a square = side² ),

Now, by the second part of the wire, a circle is formed,

if r be the radius of the circle,

Then $2\pi r = (20-x)$      ( circumference of a circle = $2\pi$ × radius ),

$\implies r =\frac{(20-x)}{2\pi }$

So,the area of second part,

$A_2=\pi (\frac{(20-x)}{2\pi })^2=\frac{(20-x)^2}{4\pi}$    ( area of a cirle = $\pi$  × radius² ),

Thus, the total sum of areas,

$A = A_1 + A_2$

$A=\frac{x^2}{16}+\frac{(20-x)^2}{4\pi}$

Differentiating with respect to x,

$A' = \frac{x}{8}-\frac{20-x}{2\pi}$

Again differentiating w. r. t. x,

$A'' = \frac{1}{8}+\frac{1}{2\pi}$

For maxima or minima,

A' = 0

$\frac{x}{8}-\frac{20-x}{2\pi}=0$

$\frac{\pi x- 80+4x}{2\pi}=0$

$\pi x - 80 + 4x=0$

$(\pi +4)x=80$

$\implies x =\frac{80}{\pi + 4}\approx 11.20$

For x = 11.20, A'' = positive,

i.e. A is minimum at x = 11.20

∵ 20 - 11.20 = 8.80

Hence, the parts must be 11.20 meters and 8.80 meters in order to minimize the sum of their areas

Answer 2

Answer:

The radius of the circle ,r= 1.43 m

The length of the side of square ,a= 2.77 m

Step-by-step explanation:

Given that

L= 20  m

Lets take radius of the circle =r m

The total parameter of the circle = 2π r

Area of circle ,A=π r²

The side of the square = a m

The total parameter of the square = 4 a

Area of square ,A'=a²

The total length ,L= 2π r+ 4 a

20  = 2π r+ 4 a

r=3.18 - 0.63 a

The total area = A+ A'

                   A"   =π r² +a²

A"= 3.14(3.18 - 0.63 a)² + a²

For minimize the area

$\dfrac{dA"}{da}=0$

3.14 x 2(3.18 - 0.63 a) (-0.63) + 2 a = 0

3.14 x (3.18 - 0.63 a) (-0.63) + a = 0

-6.21 + 1.24 a + a=0

2.24 a = 6.21

a=2.77 m

r= 3.18 - 0.63 a

r= 3.18 - 0.63 x 2.77

r=1.43 m

Therefore the radius of the circle ,r= 1.43 m

The length of the side of square ,a= 2.77 m