A wire of length 20 cm is to be cut into two pieces. One of the pieces is to be made into
a square and other into a circle. Where should the wire be cut so that the combined area
is minimum ?
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Answer:
Let one part be of length x, then the other part will be 28−x.
Let the part of the length x be covered into a circle of radius.
2πr=x
⇒r=
2π
x
Area of circle =πr
2
=π(
2π
x
)
2
=
4π
x
2
Now second part of length 280−x is covered into a square.
Side of a square =
4
28−x
Area of square =[
4
28−x
]
2
Thus total area =
4π
x
2
+[
4
28−x
]
2
dx
dA
=
4π
2x
+
16
2
(28−x)(−1)
=
2π
x
−
8
28−x
Lets take
dx
dA
=0
Thus
2π
x
−
8
28−x
=0 ....(1)
4x=28π−πx
4x+πx=28π
x[4+π]=28π
x=
4+π
28π
Other part =28−x=28−
4+π
28π
=
4+π
112+28π−28π
=
4+π
112
Now again differentiating, we get
dx
2
d
2
A
=
2π
1
+
8
1
=+ve
A is minimum.
When x=
4+π
28π
and 28−x=
4+π
112
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