Question:

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

Updated On: May 18, 2024
  • $10\, s \,cm$
  • $30\, s\, cm$
  • $20\, s\, cm$
  • $25\, s\, cm$
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The Correct Option is D

Solution and Explanation

Given that of the wire $P=20\, cm$
Then, $P=$ diameter + arc length
$20 =2 r+ S$
$S =20-2 r$
$S =2(10-r)$...(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 E (i)
$A=\frac{1}{2} \cdot 2(10-r) \cdot r$
$A=10 r-r^{2}$...(iv)
Now, $\frac{d A}{d r}=10-2 r$
For max or min area of enclosed by wire
$\Rightarrow \frac{d A}{d r}=0 \Rightarrow 10-2 r=0$
$\Rightarrow r=5$
Then, from E (iv)
$A=10(5)-(5)^{2}$
$A=50-25$
$A=25\, sq\, cm$
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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.

Read More: Application of Derivatives