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JEE Advanced 2013 Paper 1
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© examsnet.com
Question : 51
Total: 60
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:15 is converted into an open rectangular box by folding after removing squares of equal area from all four comers. If the total area of removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are
24
32
45
60
Validate
Solution:
l = 8x , b = 15x
∴ volume = (8x - 2a) (15x - 2a) (a) =
4
a
3
−
46
a
2
x
+
120
a
x
2
d
v
d
a
=
6
a
2
−
46
a
x
+
60
x
2
(
d
v
d
x
)
at
x
=
5
=
0
∴ x = 3 and
5
6
d
2
v
d
a
2
=
6
a
−
23
x
(
d
2
v
d
a
2
)
at
a
=
5
&
x
=
3
<
0
So, at x = 3 gives maxima
(
d
2
v
d
a
2
)
at
a
=
5
&
x
=
5
6
>
0
So, at x =
5
6
gives maxima
d
v
d
a
=
0
when a = 5 given (∴
4
a
2
= 100 given for maximum volume) at a = 5
by
d
v
d
a
= 0
⇒
6
x
2
- 23x + 15 = 0
x = 3 or
5
6
So by x = 3 (for max volume)
8x = 24, 15x = 45
© examsnet.com
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