A circular sector of perimeter 60 m with max | vedclass

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(B) Let $r$ be the radius and $	heta$ be the central angle in radians of the circular sector.The perimeter $P$ of the sector is given by $P = 2r + r\

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Let $r$ be the radius and $	heta$ be the central angle in radians of the circular sector.The perimeter $P$ of the sector is given by $P = 2r + r	heta = 60$.From this,we get $	heta = \frac{60 - 2r}{r}$.The area $A$ of the sector is given by $A = \frac{1}{2} r^2 	heta$.Substituting $	heta$,we get $A(r) = \frac{1}{2} r^2 \left( \frac{60 - 2r}{r} ight) = \frac{1}{2} r(60 - 2r) = 30r - r^2$.To find the maximum area,we differentiate $A(r)$ with respect to $r$ and set it to zero:$A'(r) = 30 - 2r = 0$.Solving for $r$,we get $r = 15 \ m$.Since $A''(r) = -2 < 0$,the area is maximum at $r = 15 \ m$.

$A$ circular sector of perimeter $60 \ m$ with maximum area is to be constructed. The radius of the circular arc in meters must be: (in $m$)

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