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?
Answers
Answer:
We have a piece of wire 40 centimeters 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.
Let x be the side of the square and r is the radius of the circle.
Thus, 40=perimeter of square + circumference of the circle
40
=
4
x
+
2
π
r
2
π
r
=
40
−
4
x
r
=
20
−
2
x
π
−
−
−
−
(
i
)
Now, the total area is equal to the area of the square + area of the circle
A
=
x
2
+
π
r
2
A
=
x
2
+
π
(
20
−
2
x
π
)
2
A
=
x
2
+
π
(
20
−
2
x
)
2
(
π
)
2
A
=
x
2
+
(
20
−
2
x
)
2
π
Now, on differentiating it with respect to x, we get,
A
′
=
2
x
+
2
(
20
−
2
x
)
(
−
2
)
π
A
′
=
2
x
−
4
(
20
−
2
x
)
π
On equating with zero, we get,
2
x
−
4
(
20
−
2
x
)
π
=
0
2
x
=
4
(
20
−
2
x
)
π
π
x
=
2
(
20
−
2
x
)
π
x
=
40
−
4
x
4
x
+
π
x
=
40
x
(
4
+
π
)
=
40
x
=
40
4
+
π
On again differentiating area function with respect to x, we get,
A
′′
=
2
−
4
(
−
2
)
π
A
′′
=
2
+
8
π
>
0
(
M
i
n
i
m
a
)
Since this value is always greater than zero for every value of x. Thus we will get the minimum value only.
There does not exist any maximum value.
Therefore, the area will be minimum when the side of the square is equal to
40
4
+
π
And using equation (i), the radius of the circle is
r
=
20
−
2
x
π
r
=
20
−
2
40
4
+
π
π
r
=
80
+
20
π
−
80
π
(
4
+
π
)
r
=
20
π
π
(
4
+
π
)
r
=
20
(
4
+
π
)
Thus we will get the minimum area with these dimensions.