Question

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Plzz solve question no. 32
Thanks☺☺

Answer 1

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Refer to the given attachment...


Hope it helps ☺

Answer 2

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.