Application of Derivatives Part 2

Let x + y = 36
where, x is perimeter of square and y is perimeter of circle
Then, side of square =

x

4

and radius of circle =

y

2π

Now, Sum of areas of square and circle, A=

x2

16

+

y2

4π

⇒A=

x2

16

+

(36−x)2

4π

[∵y=36−x]
For minimum area

dA

dx

=0
Now,

dA

dx

=

2x

16

+

−2(36−x)

4π

=0
⇒x=

144

π+4

Circumference of circle = y
=(36−x)
=36−

144

π+4

=

36π

π+4

According to the question,
k=

36π

π+4

⇒(

4

π

+1)k=(

4

π

+1)

36π

π+4

=36