Answer:
Therefore the circumference of the circle is 
Step-by-step explanation:
Let the side of the square be s
and the radius of the circle be r
The perimeter of the square is = 4s
The circumference of the circle is =2πr
Given that the length of the wire is 20 cm.
According to the problem,
4s + 2πr =20
⇒2s+πr =10

The area of the circle is = πr²
The area of the square is = s²
A represent the total area of the square and circle.
A=πr²+s²
Putting the value of s




For maximum or minimum 
Differentiating with respect to r

Again differentiating with respect to r
> 0
For maximum or minimum





Therefore at
, A is minimum.
Therefore the circumference of the circle is

