A wire of length 28 m is to be cut into two pieces. One of the pieces is to
be made into a square and the other into a circle. What should be the
length of the two pieces so that the combined area of the square and the
circle is minimum.
Answers
Answer:
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces, so that the combined area of the circle and the square is minimum?
Share
Study later
ANSWER
Let r be the radius of the circle and x metre be the length of each side of the square.
Then,
2πr+4x=28⇒πr+2x=14⇒r=
π
14−2x
……(i)
Let A be the combined area of the circle and the square. Then,
A=πr
2
+x
2
⇒A=π(
π
14−2x
)
2
+x
2
…..[Using (i)]
⇒A=
π
1
(14−2x)
2
+x
2
=
π
4
(7−x)
2
+x
2
⇒
dX
dA
=−
π
8
(7−x)+2x
and
dx
2
d
2
A
=
π
8
The critical numbers of A are given by
dx
dA
=0.
∴
dx
dA
=0
⇒−
π
8
(7−x)+2x=0
⇒x=
π+4
28
Clearly,
dx
2
d
2
A
=
π
8
>0 for all x.
Hence, A is minimum when x=
π+4
28
.
The lengths of two partions are 4x=
π+4
112
meter and,
28−
π+4
112
=
π+4
28π
m respectively.