Wire of length 28 m is divided into two pieces and the pieces are bent into a square and a circle

\[\text { Suppose the wire, which is to be made into a square and a circle, is cut into two pieces of length x m and y m, respectively . Then, } \]

\[x + y = 28 . . . \left( 1 \right)\]

\[\text { Perimeter of square }, 4\left( side \right) = x\]

\[ \Rightarrow \text { Side } = \frac{x}{4}\]

\[ \Rightarrow \text { Area of square } = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16}\]

\[\text { Circumference of circle }, 2\pi r = y\]

\[ \Rightarrow r = \frac{y}{2\pi}\]

\[\text { Area of circle =} \pi r^2 = \pi \left( \frac{y}{2\pi} \right)^2 = \frac{y^2}{4\pi}\]

\[\text { Now, }\]

\[z = \text { Area of square + Area of circle }\]

\[ \Rightarrow z = \frac{x^2}{16} + \frac{y^2}{4\pi}\]

\[ \Rightarrow z = \frac{x^2}{16} + \frac{\left( 28 - x \right)^2}{4\pi}\]

\[ \Rightarrow \frac{dz}{dx} = \frac{2x}{16} - \frac{2\left( 28 - x \right)}{4\pi}\]

\[\text { For maximum or minimum values of z, we must have }\]

\[\frac{dz}{dx} = 0\]

\[ \Rightarrow \frac{2x}{16} - \frac{2\left( 28 - x \right)}{4\pi} = 0 .............\left[ \text { From eq }. \left( 1 \right) \right]\]

\[ \Rightarrow \frac{x}{4} = \frac{\left( 28 - x \right)}{\pi}\]

\[ \Rightarrow \frac{x\pi}{4} + x = 28\]

\[ \Rightarrow x\left( \frac{\pi}{4} + 1 \right) = 28\]

\[ \Rightarrow x = \frac{28}{\left( \frac{\pi}{4} + 1 \right)}\]

\[ \Rightarrow x = \frac{112}{\pi + 4}\]

\[ \Rightarrow y = 28 - \frac{112}{\pi + 4} ............\left[ \text { From eq } . \left( 1 \right) \right]\]

\[ \Rightarrow y = \frac{28\pi}{\pi + 4}\]

\[ \frac{d^2 z}{d x^2} = \frac{1}{8} + \frac{1}{2\pi} > 0\]

\[\text { Thus, z is minimum when x } = \frac{112}{\pi + 4} \text { and }y = \frac{28\pi}{\pi + 4} . \]

\[\text { Hence, the length of the two pieces of wire are } \frac{112}{\pi + 4} m \text { and } \frac{28\pi}{\pi + 4} \text { m respectively }.\]

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution

Given that length of the wire is 28 m.

Let x be the length of the wire that forms the square, so, the length of second wire will be 28−x m.

As the perimeter of the square is 4×side, so,

x=4×side side= x 4

As the circumference of the circle is 2π×r, so,

28−x=2πr r= 28−x 2π

The total area of the square and circle will be,

A= ( x 4 ) 2 +π ( 28−x 2π ) 2 = x 2 16 + ( 28−x ) 2 4π

Differentiate the area with respect to x,

A ′ = 2x 16 + 2 4π ( 28−x )( −1 ) = x 8 − 1 2π ( 28−x ) (1)

Put A ′ =0,

x 8 − 1 2π ( 28−x )=0 xπ−4( 28−x ) 8π =0 x( π+4 )−112=0 x= 112 π+4

Differentiate equation (1) with respect to x,

A ″ = 1 8 + 1 2π >0

This shows that A ″ is positive, so, the value of x= 112 π+4 gives the minimum combined area of the square and the circle.

As x= 112 π+4 , so the other part will be,

28−x=28− 112 π+4 = 28π π+4

Therefore, the given combined area is minimum when the length of wire of one part is 112 π+4 cm and the length of second part is 28π π+4 cm.

Mathematics
Math Part-I - NCERT
Standard XII

Suggest Corrections

The radius of a circle is increasing uniformly at the rate of 3 cm per second. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Let r cm be the radius of the circle.From given condition, rate of increase = 3 cm per second

Let A be the area of circle,

When r = 10, rate of increase of area =

Answer

Verified

We know a square has 4 sides. Thus the perimeter of the square is the sum of the length of its side. As the length of sides of a square is the same, we can make out that \[4\times \](side of square) is equal to the length x.i.e. x is the perimeter of the square, as the wire is converted into the shape of a square.\[\therefore \] \[4\times \](side of square) = x.\[\therefore \] One side of the square\[=\dfrac{x}{4}\]\[\therefore \] Side of square\[=\dfrac{x}{4}-(1)\]The length (28 - x) m is converted into a circle. Let ‘r’ be the radius of the circle formed.As the wire is converted into a shape of the circle, the circumference of the circle becomes equal to (28 - x) m. \[\therefore \] Circumference = 28 – x.We know the circumference of a circle with radius r =\[2\pi r\].\[\begin{align} & \therefore 2\pi r=28-x \\ & \therefore r=\left( \dfrac{28-x}{2\pi } \right)m-(2) \\ \end{align}\]What we need to find is the combined length of pieces so that the combined area of circle and square is minimum.Now let us consider the total area of the circle and square as T.\[\therefore \] T = area of circle + area of square.We know the area of the circle \[=\pi {{r}^{2}}\], with radius r.Area of square \[={{\left( side \right)}^{2}}\]\[\therefore T=\pi {{r}^{2}}+{{\left( side \right)}^{2}}-(3)\]Substitute the value of r from equation (2) and side of square from equation (1) in equation (3) we get,\[T=\pi {{\left( \dfrac{28-x}{2\pi } \right)}^{2}}+{{\left( \dfrac{x}{4} \right)}^{2}}-(4)\]Now let us differentiate T with respect to x.\[\dfrac{dT}{dx}=\dfrac{d}{dx}\left[ \dfrac{\pi }{4{{\pi }^{2}}}\times {{\left( 28-x \right)}^{2}} \right]+\dfrac{d}{dx}\left( \dfrac{{{x}^{2}}}{16} \right)\]\[=\dfrac{1}{4\pi }\left[ 2\left( 28-x \right)\times \dfrac{d}{dx}\left( 28-x \right) \right]+\dfrac{2x}{16}\]\[\dfrac{dT}{dx}=\left( \dfrac{28-x}{2\pi } \right)\left( -1 \right)+\dfrac{x}{8}-(5)\]Put, \[\dfrac{dT}{dx}=0\]\[\begin{align} & \therefore \left( \dfrac{28-x}{2\pi } \right)\left( -1 \right)+\dfrac{x}{8}=0 \\ & \therefore \dfrac{x}{8}=\dfrac{28-x}{2\pi } \\ \end{align}\]Cross multiplying the above equation,\[\begin{align} & 2\pi x=8\left( 28-x \right) \\ & 2\pi x=224-8x \\ & \Rightarrow 2\pi x+8x=224 \\ \end{align}\]Now simplify the expression by dividing it by 2.\[\begin{align} & 2\pi x=8\left( 28-x \right) \\ & \pi x+4x=112 \\ & \Rightarrow x\left( \pi +4 \right)=112 \\ & x=\dfrac{112}{4+\pi }-(6) \\ \end{align}\]Now let us find, \[\dfrac{{{d}^{2}}T}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ \dfrac{-28+x}{2\pi }+\dfrac{x}{8} \right]\]\[\begin{align} & =\dfrac{1}{2\pi }\dfrac{d}{dx}\left( -28+x \right)+\dfrac{1}{8}\dfrac{d}{dx}x \\ & =\dfrac{1}{2\pi }\times 1+\dfrac{1}{8}\times 1=\dfrac{1}{2\pi }+\dfrac{1}{8} \\ \end{align}\]\[\therefore \dfrac{{{d}^{2}}T}{d{{x}^{2}}}=\dfrac{1}{2\pi }+\dfrac{1}{8}\], which is greater than zero.Hence, \[\dfrac{{{d}^{2}}T}{d{{x}^{2}}}>0\] at \[x=\dfrac{112}{4+\pi }\]. Hence, we can say that at \[x=\dfrac{112}{4+\pi }\] total area is minimum.Now let us find the length of the other part.Length of other part =28 – x\[=28-\dfrac{112}{4+\pi }=\dfrac{28\left( 4+\pi \right)-112}{\left( 4+\pi \right)}\]Simplify the expression,\[\begin{align} & =\dfrac{112+28\pi -112}{4+\pi } \\ & =\dfrac{28\pi }{4+\pi } \\ \end{align}\]\[\therefore \]Length of 1st part\[=\dfrac{112}{4+\pi }\]Length of 2nd part\[=\dfrac{28\pi }{4+\pi }\]Note:Now let us consider the first part of length ‘x’ as circumference of circle and (28 - x) as the perimeter of the square.The radius of circle, \[2\pi r=x\Rightarrow r=\dfrac{x}{2\pi }\]Side of the square\[=\dfrac{side}{4}=\dfrac{28-x}{4}\]Total area, \[T={{\left( \dfrac{28-x}{4} \right)}^{2}}+{{\left( \dfrac{x}{2\pi } \right)}^{2}}\times \pi \]Differentiate T w.r.t to x.\[\begin{align} & \dfrac{dT}{dx}=\dfrac{1}{16}\times 2\left( 28-x \right)\left( -1 \right)+\dfrac{\pi }{4{{\pi }^{2}}}\times 2x \\ & =\dfrac{-\left( 28-x \right)}{8}+\dfrac{x}{2\pi } \\ \end{align}\]Put, \[\dfrac{dT}{dx}=0\]\[\Rightarrow \dfrac{28-x}{8}=\dfrac{x}{2\pi }\]Cross multiplying and solving, we get,\[\begin{align} & 2\pi \left( 28-x \right)=8x \\ & 56\pi -2\pi x=8x \\ & \Rightarrow 28\pi -\pi x=4x \\ & \Rightarrow x\left( 4+\pi \right)=28\pi \\ & \therefore x=\dfrac{28\pi }{4+\pi } \\ \end{align}\]Taking, \[\dfrac{{{d}^{2}}T}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{x-28}{8}+\dfrac{x}{2\pi } \right)=\dfrac{1}{8}+\dfrac{1}{2\pi }>0\]Length of 1st part, \[x=\dfrac{28\pi }{4+\pi }\]Length of 2nd part, \[x=28-x=28-\dfrac{28\pi }{4+\pi }=\dfrac{112+28\pi -28\pi }{4+\pi }=\dfrac{112}{4+\pi }\]\[\therefore \] on solving answers are the same.Here, the area is minimum at \[x=\dfrac{28\pi }{4+\pi }\].

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