Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

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

$\Rightarrow \; b²\; = \; 4\; a²\; - <em>l</em>²$

$$\begin{gathered} \Rightarrow \mathrm{b} = \sqrt{ 4 {\mathrm{a}}^2 -\mathit{ }{l}^2} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{rectangle}, \mathrm{A} =\mathit{ }l \sqrt{ 4 {\mathrm{a}}^2 - l^2} \\ \therefore \frac{\mathrm{dA}}{\mathrm{dl}} = \sqrt{ 4 {\mathrm{a}}^2 - l^2} + \mathit{ }l\mathit{ }\frac{1}{2 \sqrt{ 4 {\mathrm{a}}^2 -\mathit{ }{l}^2}} \left( - 2 l\mathit{ }\right) = \sqrt{ 4 {\mathrm{a}}^2 - l^2} - \frac{l^2}{ \sqrt{ 4 {\mathrm{a}}^2 -\mathit{ }{l}^2}} \\ = \frac{ 4 {\mathrm{a}}^2 - 2 l^2}{\sqrt{ 4 {\mathrm{a}}^2 - l^2}} \\ \frac{{\mathrm{dA}}^2}{{\mathrm{dl}}^2} = \frac{ \sqrt{ 4 {\mathrm{a}}^2 -\mathit{ }{l}^{2 }} \left( - 4 l \right) - \left( 4 {\mathrm{a}}^2 - 2 l^2 \right) {\displaystyle \frac{\left( - 2\mathit{ }l \right)}{2 \sqrt{ 4 {\mathrm{a}}^2 -\mathit{ }{l}^2}}}}{\left(4 {\mathrm{a}}^2 - \mathit{ }{l}^2 \right)} \\ = \frac{\left(4 {\mathrm{a}}^2 - \mathit{ }{l}^2 \right) \left( - 4\mathit{ }l \right) + l\mathit{ }\left( 4 {\mathrm{a}}^2 - 2\mathit{ }{l}^2 \right)}{{\left(4 {\mathrm{a}}^2 - l^2 \right)}^{{\displaystyle \frac{3}{2}}}} \\ = \frac{ - 12 {\mathrm{a}}^2 l\mathit{ }+ 2\mathit{ }{l}^3}{{\left(4 {\mathrm{a}}^2 - \mathit{ }{l}^2 \right)}^{\frac{3}{2}}} = \frac{- 2\mathit{ }l \left( 6 {\mathrm{a}}^2 - l^2 \right)}{{\left(4 {\mathrm{a}}^2 - l^2 \right)}^{\frac{3}{2}}} \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \frac{\mathrm{dA}}{\mathrm{d}l } = 0 \mathrm{gives} 4 {\mathrm{a}}^2 = 2 l^2 \Rightarrow \mathit{ }\mathit{ }l\mathit{ }= \sqrt{ 2 \mathrm{a}} \\ \Rightarrow \mathrm{b} = \sqrt{4 {\mathrm{a}}^2 - 2 {\mathrm{a}}^2 } = \sqrt{ 2 {\mathrm{a}}^2} = \sqrt{ 2} \mathrm{a} \\ \mathrm{When}\mathit{ }\mathit{ }l = \sqrt{ 2} \mathrm{a}, \\ \frac{{\mathrm{dA}}^2}{\mathrm{d}{l}^2 } = \frac{- 2 \left(\sqrt{ 2} \mathrm{a} \right) \left( 6 {\mathrm{a}}^2 - 2 {\mathrm{a}}^2 \right)}{2 \sqrt{ 2} {\mathrm{a}}^3} = \frac{- 8 \sqrt{ 2} {\mathrm{a}}^3}{2 \sqrt{ 2} {\mathrm{a}}^3} = - 4 < 0 \end{gathered}$$

$\therefore$ Thus, frrom the second derivative test, when  l = $\sqrt{2} a$, the area of the rectangle is maximum.

Since  l = b = $\sqrt{2} \mathrm{a}$,  the rectangle is square.

Hence, of all the rectangles inscribed in the given circle, the square has the maximum area.