Let the rectangle of length l and breadth b be inscribed in circle of radius a.

Then, the diagonal of the rectangle passes through the centre and is of length 2a cm.
Now, by applying the Pythagoras Theorem, we have:
(2a)2 = l2+b2
⇒ b2 = 4a2−l2
⇒ b = √4a2−l2
∴ Area of rectangle , A = lb = r √4a2−l2
∴

dA

dl

= √4a2−l2 + l .

1

2√4a2−l2

(- 2l) = √4a2−l2 -

l2

√4a2−l2

=

4a2−2l2

√4a2−l2

d2A

dl2

=
=

(4a2−l2)(−4l)+l(4a2−2l2)

(4a2−l2) 3 2

=

−12a2l+2l3

(4a2−l2) 3 2

=

−2l(6a2−l2)

(4a2−l2) 3 2

Now,

dA

dl

= 0 gives 4a2 = 2l2 ⇒ l = √2 a
when l = √2 a

d2A

dl2

=

−2(√2a)(6a2−2a2)

2√2a3

=

−8√2a3

2√2a3

= - 4 < 0
∴ Thus, from the second derivative test, when l = √2 a , the area of the rectangle is maximum.
Since l = b = √2 a , the rectangle is a square