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Plzz solve question no. 32
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Answers
Answered by
10
Hii There!!!
Refer to the given attachment...
Hope it helps ☺
Refer to the given attachment...
Hope it helps ☺
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Khushi0511:
Ok you tell me Msg kyun krna hai? Any reason...??
becoz, I'm in 11th standard nd tumhare itne danger ques mein solve nhi kr pati hun aaj ka ex. le lo
Sis...Not madam..bro.. : )
Yup bro... :-)
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Answered by
1
Answer:
Let x,y be the length and breadth of rectangle whose are is A and perimeter is P.
⇒ P=2(x+y) --- ( 1 )
We know, A=x×y
∴ y=
x
A
---- ( 2 )
Substituting value of ( 2 ) in ( 1 ) we get,
⇒ P=2(x+
x
A
)
For maximum or minimum values of perimeter P
⇒
dx
dP
=2(1−
x
2
A
)=0
⇒ 1−
x
2
A
=0
⇒ x
2
=A
⇒ x=
A
[ Dimension of rectangle is always positive ]
Now,
dx
2
d
2
P
=2(0−A×
x
3
(−1)
)=
x
3
2A
∴ [
dx
2
d
2
P
]
x=
A
=
(
A
)
3
2a
>0
i.e., for x=
A
,P (perimeter of rectangle ) is smallest.
y=
x
A
=
A
A
=
A
Hence, for smallest perimeter, length and breadth of rectangle are equal (x=y=
A
) i.e., rectangle is square.
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