Question

A piece of wire 40 cm long is cut into two pieces. One piece is bent into the shape of a square and the other is bent into the shape of a circle. How should the wire be cut so that the total area enclosed is a (a) maximum and (b) minimum?
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Answer 1

Answer:

ANSWER

Let the 2 pieces be x,y

Given,

x+y=20

Perimeter of square =x

Side=

4

x

Area of square =

16

x

2

Perimeter of triangle =y

Side=

3

y

Area of triangle =

4

3

(

3

y

)

2

=

36

3

y

2

z= area of square + area of triangle

z=

16

x

2

+

36

3

y

2

z=

16

x

2

+

36

3

(20−x)

2

differentiating the equation, we get,

dx

dz

=

8

x

−

18

3

(20−x)

equate,

dx

dz

=0

we have,

8

x

−

18

3

(20−x)

=0

8

x

=

18

3

(20−x)

solving the above equation, we get,

→x=

(9+4

3

)

80

3

y=20−

(9+4

3

)

80

3

→y=

(9+4

3

)

180

dx

2

d

2

z

=

8

1

+

18

3

>0

Thus z is minimum, when x=

(9+4

3

)

80

3

and y=

(9+4

3

)

180

Hence, wire of length 20cm should be cut into pieces of lengths

(9+4

3

)

80

3

,

(9+4

3

)

180

m