In $\Delta$DBF $\tan 60^\circ = {{2b} \over {2\sqrt 2 - l}} \Rightarrow b = {{\sqrt 3 \left( {2\sqrt 2 - l} \right)} \over 2}$ A = Area of rectangle = l $\times$ b $A = l \times {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right)$ ${{dA} \over {dl}} = {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right) - {{l.\sqrt 3 } \over 2} = 0$ $l = \sqrt 2$ $A = l \times b = \sqrt 2 \times {{\sqrt 3 } \over 2}\left( {\sqrt 2 } \right) = \sqrt 3$ $\Rightarrow$ A 2 = 3