Application of Derivatives Part 2

Let two pieces of wire one of length x and other of the length20 − x.
Wire of length x is made into a square ∴ Area of square =(

x

4

)2=AS(let)
Wire of length (20 − x) is made into a regular hexagon. Area of hexagon =6×

√3

4

a2
(Let)AH=

3√3

2

(

20−x

6

)2 [∵a=

20−x

6

]
Sum of both area
A=AS+AH
=

x2

16

+

√3

24

(20−x)2

dA

dx

=

x

8

−

√3

12

(20−x)=

3x−40√3+2√3x

24

dA

dx

=0
⇒x=

40√3

3+2√3

=

40

√3+2

=40(2−√3)

d2A

dx2

=

3+2√3

24

>0
⇒ Area will be minimum, when x=40(2−√3)
∴ Side of hexagon =

20−40(2−√3)

6

=

20√3−30

3

=

10(2√3−3)

3

=

10

2√3+3