Respuesta :
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 Â m
Lets take radius of the circle =r m
The total parameter of the circle = 2Ï€ 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= 2Ï€ r+ 4 a
20 Â = 2Ï€ 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
[tex]\dfrac{dA"}{da}=0[/tex]
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
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 ),
[tex]\implies a =\frac{x}{4}[/tex]
So, the area of first part,
[tex]A_1=(\frac{x}{4})^2=\frac{x^2}{16}[/tex]   ( 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 [tex]2\pi r = (20-x)[/tex]    ( circumference of a circle = [tex]2\pi[/tex] × radius ),
[tex]\implies r =\frac{(20-x)}{2\pi }[/tex]
So,the area of second part,
[tex]A_2=\pi (\frac{(20-x)}{2\pi })^2=\frac{(20-x)^2}{4\pi}[/tex]   ( area of a cirle = [tex]\pi[/tex]  × radius² ),
Thus, the total sum of areas,
[tex]A = A_1 + A_2[/tex]
[tex]A=\frac{x^2}{16}+\frac{(20-x)^2}{4\pi}[/tex]
Differentiating with respect to x,
[tex]A' = \frac{x}{8}-\frac{20-x}{2\pi}[/tex]
Again differentiating w. r. t. x,
[tex]A'' = \frac{1}{8}+\frac{1}{2\pi}[/tex]
For maxima or minima,
A' = 0
[tex]\frac{x}{8}-\frac{20-x}{2\pi}=0[/tex]
[tex]\frac{\pi x- 80+4x}{2\pi}=0[/tex]
[tex]\pi x - 80 + 4x=0[/tex]
[tex](\pi +4)x=80[/tex]
[tex]\implies x =\frac{80}{\pi + 4}\approx 11.20[/tex]
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
