Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

A wire

10

metres long is cut into two parts. One part is bent into the shape of a circle and the other into the shape of an equilateral triangle. How should the wire be cut so that the combined area of the two figures is as small as possible.

Accepted Answer

Let $r$ meter be the radius of circle and $a$ meter be the side of equilateral triangle.

Then, $2 π r + 3 a = 10$ $. . . . . . . . . . . . . . . . . . . (1)$

Let, $A =$ combined area of equilateral triangle and circle

$\Rightarrow A = π r^{2} + \frac{\sqrt{3}}{4} a^{2}$

$A = {πr}^{2} + \frac{\sqrt{3}}{4} {(\frac{10 - 2 πr}{3})}^{2}$ [using $(1)$ ]

On differentiating both the sides w.r.t. $x$ , we get

$\frac{d A}{d r} = 2 πr + \frac{\sqrt{3}}{36} 2 (10 - 2 πr) (- 2 π)$

$\frac{d A}{d r} = 2 πr - \frac{\sqrt{3}}{9} π (10 - 2 πr)$

and $\frac{d^{2} A}{d r^{2}} = 2 π - \frac{\sqrt{3} π}{9} (0 - 2 π)$

$\frac{d^{2} A}{d r^{2}} = 2 π + \frac{2 \sqrt{3}}{9} π^{2}$

Now, For maxima/minima, $\frac{d A}{d r} = 0$

$\Rightarrow 18 πr - 10 \sqrt{3} π + 2 \sqrt{3} π^{2} r = 0$

$\Rightarrow (18 + 2 \sqrt{3} π) r = 10 \sqrt{3}$

$\Rightarrow r = \frac{10 \sqrt{3}}{(18 + 2 \sqrt{3} π)}$

$\Rightarrow r = \frac{10 \sqrt{3}}{2 \sqrt{3} (3 \sqrt{3} + π)}$

$\Rightarrow r = \frac{5}{(3 \sqrt{3} + π)}$

Now, at $r = \frac{5}{(3 \sqrt{3} + π)}$ ;

$(\frac{d^{2} A}{d r^{2}}) = 2 π + \frac{2 \sqrt{3}}{9} π^{2} (∵ 2 π + \frac{2 \sqrt{3}}{9} π^{2} > 0)$

Thus, $r = \frac{5}{(3 \sqrt{3} + π)}$ is the point of minima.

Therefore, area $A$ is minimum when $r = \frac{5}{(3 \sqrt{3} + π)}$ .

Thus length of piece of wire bent into the form of circle $= 2 πr = \frac{10 π}{(3 \sqrt{3} + π)} m$

and length of piece of wire bent into to form equilateral triangle $= 3 a = 10 - \frac{10 π}{(3 \sqrt{3} + π)} = \frac{30 \sqrt{3}}{(3 \sqrt{3} + π)} m$

A wire $10$ metres long is cut into two parts. One part is bent into the shape of a circle and the other into the shape of an equilateral triangle. How should the wire be cut so that the combined area of the two figures is as small as possible.

Important Questions on Applications of Derivatives

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