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: $(2a)^2$ = $l^2+b^2$ ⇒ $b^2$ = $4a^2-l^2$ ⇒ b = $\sqrt{4a^2-l^2}$ ∴ Area of rectangle , A = lb = r $\sqrt{4a^2-l^2}$ ∴ $\frac{dA}{dl}$ = $\sqrt{4a^2-l^2}$ + l . $\frac{1}{2\sqrt{4a^2-l^2}}$ (- 2l) = $\sqrt{4a^2-l^2}$ - $\frac{l^2}{\sqrt{4a^2-l^2}}$ = $\frac{4a^2-2l^2}{\sqrt{4a^2-l^2}}$ $\frac{d^2A}{dl^2}$ = = $\frac{(4a^2-l^2)(-4l)+l(4a^2-2l^2)}{(4a^2-l^2)^{3/2}}$ = $\frac{-12a^2l+2l^3}{(4a^2-l^2)^{3/2}}$ = $\frac{-2l(6a^2-l^2)}{(4a^2-l^2)^{3/2}}$ Now, $\frac{dA}{dl}$ = 0 gives $4a^2$ = $2l^2$ ⇒ l = $\sqrt{2}$ a when l = $\sqrt{2}$ a $\frac{d^2A}{dl^2}$ = $\frac{-2(\sqrt{2}a)(6a^2-2a^2)}{2\sqrt{2}a^3}$ = $\frac{-8\sqrt{2}a^3}{2\sqrt{2}a^3}$ = - 4 < 0 ∴ Thus, from the second derivative test, when l = $\sqrt{2}$ a , the area of the rectangle is maximum. Since l = b = $\sqrt{2}$ a , the rectangle is a square