Question:

A circular sector of perimeter $60$ metre with maximum area is to be constructed. The radius of the circular arc in metre must be

Updated On: Sep 4, 2024
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The Correct Option is B

Solution and Explanation

Perimeter of sector = $2r + r \theta$
$\Rightarrow \, 60 = 2r + r\theta$ (given)

$\Rightarrow \; \theta = \frac{60 - 2r}{r}$
Area of sector, $A = \frac{\pi r^2 \theta}{360^{\circ}}$
$= \frac{\pi r^{2} \left(60 -2r\right)}{r 360} $
$ = \frac{\pi r}{180} \left(30 -r\right) $
$\Rightarrow \frac{dA}{dr} = \frac{\pi}{180} \left(30 -2r\right) $
For maximum area, $ \frac{dA}{dr} = 0 $
$ \Rightarrow 30-2r=0 $
$ \Rightarrow r = 15 $
$\therefore \frac{d^{2}A}{dr^{2}} = \frac{\pi}{180} \left(0-2\right) = \frac{-\pi}{90}
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

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