A piece of wire 40 cm long is cut into two pieces. One piece is bent into the shape of a square and the other is bent into the shape of a circle. How should the wire be cut so that the total area enclosed is a (a) maximum and (b) minimum?
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ANSWER
Let the 2 pieces be x,y
Given,
x+y=20
Perimeter of square =x
Side=
4
x
Area of square =
16
x
2
Perimeter of triangle =y
Side=
3
y
Area of triangle =
4
3
(
3
y
)
2
=
36
3
y
2
z= area of square + area of triangle
z=
16
x
2
+
36
3
y
2
z=
16
x
2
+
36
3
(20−x)
2
differentiating the equation, we get,
dx
dz
=
8
x
−
18
3
(20−x)
equate,
dx
dz
=0
we have,
8
x
−
18
3
(20−x)
=0
8
x
=
18
3
(20−x)
solving the above equation, we get,
→x=
(9+4
3
)
80
3
y=20−
(9+4
3
)
80
3
→y=
(9+4
3
)
180
dx
2
d
2
z
=
8
1
+
18
3
>0
Thus z is minimum, when x=
(9+4
3
)
80
3
and y=
(9+4
3
)
180
Hence, wire of length 20cm should be cut into pieces of lengths
(9+4
3
)
80
3
,
(9+4
3
)
180
m
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