answers: 1

Register to add an answer

The time for answering the question is over

Answer:

Answer:

Let one part be of length x, then the other part will be 28−x.

Let the part of the length x be covered into a circle of radius.

2πr=x

⇒r=

2π

x

Area of circle =πr

2

=π(

2π

x

)

2

=

4π

x

2

Now second part of length 280−x is covered into a square.

Side of a square =

4

28−x

Area of square =[

4

28−x

]

2

Thus total area =

4π

x

2

+[

4

28−x

]

2

dx

dA

=

4π

2x

+

16

2

(28−x)(−1)

=

2π

x

−

8

28−x

Lets take

dx

dA

=0

Thus

2π

x

−

8

28−x

=0 ....(1)

4x=28π−πx

4x+πx=28π

x[4+π]=28π

x=

4+π

28π

Other part =28−x=28−

4+π

28π

=

4+π

112+28π−28π

=

4+π

112

Now again differentiating, we get

dx

2

d

2

A

=

2π

1

+

8

1

=+ve

A is minimum.

When x=

4+π

28π

and 28−x=

4+π

112

244

vague

For answers need to register.

Contacts

mail@expertinstudy.com

Feedback

Expert in study

About us

For new users

For new experts

Terms and Conditions