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JEE Main Math Class 12 Application of Derivatives Part 3 Questions
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© examsnet.com
Question : 3
Total: 40
Let a rectangle
A
B
C
D
of sides 2 and 4 be inscribed in another rectangle
P
Q
R
S
such that the vertices of the rectangle
A
B
C
D
lie on the sides of the rectangle
PQRS
. Let
a
and
b
be the sides of the rectangle
PQRS
when its area is maximum. Then
(
a
+
b
)
2
is equal to :
[5 Apr 2024 Shift 1]
72
60
80
64
Validate
Solution:
👈: Video Solution
Area
=
(
4
cos
θ
+
2
s
i
n
θ
)
(
2
cos
θ
+
4
s
i
n
θ
)
=
8
cos
2
θ
+
16
s
i
n
θ
cos
θ
+
4
s
i
n
θ
cos
θ
+
8
s
i
n
2
θ
=
8
+
20
s
i
n
θ
cos
θ
=
8
+
10
s
i
n
2
θ
Max Area
=
8
+
10
=
18
(
s
i
n
2
θ
=
1
)
θ
=
45
∘
(
a
+
b
)
2
=
(
4
cos
θ
+
2
s
i
n
θ
+
2
cos
θ
+
4
s
i
n
θ
)
2
=
(
6
cos
θ
+
6
s
i
n
θ
)
2
=
36
(
s
i
n
θ
+
cos
θ
)
2
=
36
(
√
2
)
2
=
72
© examsnet.com
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