Question

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in $s m$) of the flower-bed, is :

• A. 10
• B. 25
• C. 30
• D. 12.5

Answer 1

Answer: (B)

25

Answer 2

$2r + \theta r = 20$...(i)
$A =$ area = $\frac{\theta}{2\pi}\times\pi r^{2} = \frac{\theta r^{2}}{2}$...(ii)
$A = \frac{r^{2}}{2}\left(\frac{20-2r}{r}\right)$
$A = \left(\frac{20r-2r^{2}}{2}\right) = 10r - r^{2}$
A to be maximum
$\frac{dA}{dr} = 10-2r = 0 \Rightarrow r = 5$
$\frac{d^{2}A}{dr^{2}} = -2<0$
Hence for $r = 5,$ A is maximum
Now, $10 + \theta\cdot5 = 20\Rightarrow \theta = 2$ (radian)
Area $= \frac{2}{2\pi}\times\pi\left(5\right)^{2} = 25\, sq\, m$