CGPET 2022 Solved Paper

Let the length of the side of the square be x m and the radius of the circle be y metres. It is given that the length of the wire is 28m
∴4x+2πy=28
⇒2x+πy=14 ...(i)
Let A be the combined area of the square and circle. Then,
A=x2+πy2 ...(ii)
⇒A=x2+π(

14−2x

π

)2 ...[Using‌Eq(i)]
=x2+

1

π

(14−2x)2
⇒

dA

dx

=2x+

2

π

(14−2x)(−2)
=2x−

4

π

(14−2x)
and

d2A

dx2

=2−

4

π

(−2)=2+

8

π

The critical numbers of A are given by

dA

dx

=0
∴2x−

4

π

(14−2x)=0
⇒x=

28

π+4

Clearly,
(

d2A

dx2

)x=

28

π+4

=2+

8

π

>0
Thus, A is minimum when
x=

28

π+4

. Putting x=

28

π+4

in Eq.(i) we get
y=

14

π+4

So, length of the two piece of wire are
4x=4×

28

4+π

=

112

4+π

m
and
2πy=2π×

14

π+4

=

28π

π+4

m