Official solution
We have
Total length = r + r + r\[\theta \] = 20
\[ \Rightarrow \] 2r + r\[\theta \] = 20
\[ \Rightarrow \] \[\theta = {{20 - 2r} \over r}\] .......(1)
A = Area = \[{\theta \over {2\pi }} \times \pi {r^2}\] = \[{1 \over 2}{r^2}\theta \] = \[{1 \over 2}{r^2}\left( {{{20 - 2r} \over r}} \right)\]
\[ \Rightarrow \] A = 10r -- r2
For A to be maximum
\[{{dA} \over {dr}} = 0\]
\[ \Rightarrow \] 10 -- 2r = 0
\[ \Rightarrow \] r = 5
\[{{{d^2}A} \over {d{r^2}}} = - 2 < 0\]
\[ \therefore \] For r = 5, A is maximum
From (1)
\[{\theta = {{20 - 2\left( 5 \right)} \over r}}\] = 2
A = \[{2 \over {2\pi }} \times \pi {\left( 5 \right)^2}\] = 25 sq. m