Answer:
Option C
Explanation:
Total length = 2r+rθ =20
$\Rightarrow \theta =\frac{20-2r}{r}$
Now, area of flower-bed,
$A= \frac{1}{2}r^{2}\theta \Rightarrow A =\frac{1}{2}r^{2}(\frac{20-2r)}{r}$
$\Rightarrow A=10r-r^{2}$
$\therefore \frac{\text{d}A}{\text{d}r}=10-2r$
For maxima or minima , put $\therefore \frac{\text{d}A}{\text{d}r}=0$
$\therefore 10-2r=0\Rightarrow r=5$
$\therefore A_{max}= \frac{1}{2}(5)^{2}[\frac{20-2(5)}{5}]$
$ = \frac{1}{2}\times25\times2=25 sq.m$