mal Each of the small piece bent into a square of side 2 cm.
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Answered by
0
ANSWER
Length of wire =50 m (Given)
Let length of one piece for shape of square =x m
∴ Length of other piece for shape of circle
=(50−x)m
Now perimeter of square =4a=x
⇒a=
4
x
and circumference of circle =2πr=50−x
⇒r=
2π
50−x
Combined Area =a
2
+πr
2
=
16
x
2
+π⋅(
2π
50−x
)
2
=
16
x
2
+π⋅
4π
2
(50−x)
2
A=
16
x
2
+
4π
(50−x)
2
Differentiating w.r. to x, we get
dx
dA
=
16
2x
+
4π
2(50−x)(−1)
dx
dA
=
8
x
+
2π
(x−50)
=
8π
πx+4x−200
=
8π
x(4+π)−200
For extremum,
dx
dA
=0
∴x(4+π)−200=0
x=
4+π
200
dx
2
d
2
A
>0
A is minimum at x=
4+π
200
∴ Length of square of square wire, x=
4+π
200
m
and length of circle wire =50−x
=50−
4+π
200
=
4+π
50π
m
Length of wire =50 m (Given)
Let length of one piece for shape of square =x m
∴ Length of other piece for shape of circle
=(50−x)m
Now perimeter of square =4a=x
⇒a=
4
x
and circumference of circle =2πr=50−x
⇒r=
2π
50−x
Combined Area =a
2
+πr
2
=
16
x
2
+π⋅(
2π
50−x
)
2
=
16
x
2
+π⋅
4π
2
(50−x)
2
A=
16
x
2
+
4π
(50−x)
2
Differentiating w.r. to x, we get
dx
dA
=
16
2x
+
4π
2(50−x)(−1)
dx
dA
=
8
x
+
2π
(x−50)
=
8π
πx+4x−200
=
8π
x(4+π)−200
For extremum,
dx
dA
=0
∴x(4+π)−200=0
x=
4+π
200
dx
2
d
2
A
>0
A is minimum at x=
4+π
200
∴ Length of square of square wire, x=
4+π
200
m
and length of circle wire =50−x
=50−
4+π
200
=
4+π
50π
m
Answered by
0
Answer:
its area is 4cm^2 and its perimeter is 8cm
Step-by-step explanation:
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