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JEE Main Math Class 12 Application of Derivatives Part 2 Questions
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© examsnet.com
Question : 61
Total: 100
A wire of length
36
m
is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum and the circumference of the circle is
k
(
m
)
, then
(
4
π
+
1
)
k
is equal to
[26 Aug 2021 Shift 1]
Your Answer:
Validate
Solution:
Let x + y = 36
where, x is perimeter of square and y is perimeter of circle
Then, side of square
=
x
4
and radius of circle
=
y
2
π
Now, Sum of areas of square and circle,
A
=
x
2
16
+
y
2
4
π
⇒
A
=
x
2
16
+
(
36
−
x
)
2
4
π
[
∵
y
=
36
−
x
]
For minimum area
dA
dx
=
0
Now,
dA
dx
=
2
x
16
+
−
2
(
36
−
x
)
4
π
=
0
⇒
x
=
144
π
+
4
Circumference of circle = y
=
(
36
−
x
)
=
36
−
144
π
+
4
=
36
π
π
+
4
According to the question,
k
=
36
π
π
+
4
⇒
(
4
π
+
1
)
k
=
(
4
π
+
1
)
36
π
π
+
4
=
36
© examsnet.com
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