The perimeter of a sector is p. The area of t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $r$ be the radius and $s$ be the arc length of the sector. The perimeter $p$ is given by $p = 2r + s$,so $s = p - 2r$.The area $A$ of the sect

Vedclass · Accepted Answer

$\frac{p}{4}$

Vedclass · Answer

(D) Let $r$ be the radius and $s$ be the arc length of the sector. The perimeter $p$ is given by $p = 2r + s$,so $s = p - 2r$.The area $A$ of the sector is given by $A = \frac{1}{2}rs$.Substituting $s = p - 2r$ into the area formula,we get $A = \frac{1}{2}r(p - 2r) = \frac{1}{2}pr - r^2$.To find the maximum area,we differentiate $A$ with respect to $r$ and set the derivative to zero:$\frac{dA}{dr} = \frac{1}{2}p - 2r = 0$.Solving for $r$,we get $2r = \frac{p}{2}$,which implies $r = \frac{p}{4}$.Since $\frac{d^2A}{dr^2} = -2 < 0$,the area is maximum at $r = \frac{p}{4}$.

The perimeter of a sector is $p$. The area of the sector is maximum when its radius is

Similar Questions

Two particles move in the same straight line starting at the same moment from the same point in the same direction. The first moves with constant velocity $u$ and the second starts from rest with constant acceleration $f$. Then,

The maximum value of the function $f(x) = \frac{\log x}{x}, x > 0$ is

Let $AD$ and $BC$ be two vertical poles at $A$ and $B$ respectively on a horizontal ground. If $AD = 8 \ m$,$BC = 11 \ m$ and $AB = 10 \ m$,then the distance (in meters) of point $M$ on $AB$ from the point $A$ such that $MD^2 + MC^2$ is minimum,is

The condition for the function $f(x) = x^3 + px^2 + qx + r$ $(x \in R)$ to have no extreme value is:

Prove that the function $h(x) = x^{3} + x^{2} + x + 1$ does not have any local maxima or local minima.

Vedclass Test Series

Exam Paper Generator

Online Exam Module