20 m of wire is available to fence a flowerb | vedclass

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

Question

(D) Let $r$ be the radius of the circular sector and $l$ be the length of the arc. The perimeter of the sector is given by $P = 2r + l = 20 \ m$.There

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Let $r$ be the radius of the circular sector and $l$ be the length of the arc. The perimeter of the sector is given by $P = 2r + l = 20 \ m$.Therefore,the arc length is $l = 20 - 2r$.The area of a circular sector is given by $A = \frac{1}{2} r l$.Substituting the value of $l$,we get $A(r) = \frac{1}{2} r (20 - 2r) = 10r - r^2$.To find the maximum area,we differentiate $A(r)$ with respect to $r$ and set it to zero:$\frac{dA}{dr} = 10 - 2r$.Setting $\frac{dA}{dr} = 0$,we get $10 - 2r = 0$,which implies $r = 5 \ m$.To verify,the second derivative is $\frac{d^2A}{dr^2} = -2$,which is less than $0$,confirming that the area is maximum at $r = 5 \ m$.

$20 \ m$ of wire is available to fence a flowerbed in the form of a circular sector. If the flowerbed is to have maximum surface area,then the radius of the circle is (in $m$)

Similar Questions

The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to

At what value of $x$ does the function $f(x) = \sin x(1 + \cos x)$ have a maximum value?

The minimum value of the expression $7 - 20x + 11x^2$ is (in $/11$)

The maximum value of $xy$ subject to $x+y=7$ is

The minimum value of the function $f(x) = 2 x^3 - 15 x^2 + 36 x - 48$ on the set $A = \{x \mid x^2 + 20 \le 9 x\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module