Question

A wire of length $20cm$ is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is ____

• A. $10sq.cm$
• B. $30sq.cm$
• C. $20sq.cm$
• D. $25sq.cm$

Answer 1

Answer: (D)

Given that of the wire $P=20cm$
Then, $P=$ diameter + arc length
$20=2r+S$
$S=20-2r$
$S=2(10-r)$...(i)
Also, know that area of semicircle
$A=\frac{1}{2}\pi r^2$...(ii)
$\Rightarrow A=\frac{1}{2}(\pi r)(r)$
$\because$ Angle $=\frac{\text{ Arc }}{\text{ Radius }}$
$\Rightarrow \pi =\frac{S}{r}$
$\Rightarrow S=r\pi$ for straight length of wire
$\Rightarrow A=\frac{1}{2}S\cdot r$...(iii)
From Eq. (i)
$A=\frac{1}{2}\cdot 2(10-r)\cdot r$
$A=10r-r^2$...(iv)
Now, $\frac{dA}{dr}=10-2r$
For max or min area of enclosed by wire
$\Rightarrow \frac{dA}{dr}=0\Rightarrow 10-2r=0$
$\Rightarrow r=5$
Then, from Eq. (iv)
$A=10(5)-(5{)}^2$
$A=50-25$
$A=25sqcm$