(a) Length of the wire is l = 20 cm
Given that wire is formed in to the shade of the sector of circle or radius r cm and angle θ.
l = 2r + \(\frac{2\pi r}{2\pi}\times\theta\)
l = 2r + rθ
2r + rθ = 20 (\(\because\) l = 20 cm)
rθ = \(\frac{20-2r}r\)---(1)
Area of sector = \(\frac{\pi r^2}{2\pi}\times\theta\) = \(\frac{r^2}2\times\frac{20-2r}r\) (From (1))
= \(\frac r2\times2(10-r)\)
= r (10 - r) cm2
(b) Area of sector is A = r (10 - r) = 10r - r2
\(\therefore\) \(\frac{dA}{dr}\) = 10 - 2r
\(\frac{dA}{dr}\) = 0 gives r = 5 \(\therefore\) Critical point is r = 5
& \(\frac{d^2A}{dr^2} = -2 < 0\)
\(\therefore\) Critical point r = 5 is local maximum of function A.
\(\therefore\) Maximum area = 5(10 -5) = 5 x 5 = 25 cm2