# A wire of length 28m 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?

Last updated date: 19th Mar 2023

•

Total views: 303.6k

•

Views today: 6.83k

Answer

Verified

303.6k+ views

Hint: Put the length of the first part as ‘x’ and 2nd part as (28 - x). Take x as the perimeter of the square and (28 - x) as the perimeter of the circle. Find the side of the square and radius of the square. Then find the combined length of square and radius such that their area is minimum.

Complete step-by-step answer:

It’s said that a wire of 28m length is cut into 2 parts. We don’t know the length at which it was cut. So, let us consider the length of the first part as ‘x’.

So, the length of the second part will be (28 - x), as 28m is the total length of wire.

It’s said that one part of the wire is converted into a square and the other part is converted into a circle.

Let us consider that the length x m is converted into a square.

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 \[x=\dfrac{28\pi }{4+\pi }\].

Complete step-by-step answer:

It’s said that a wire of 28m length is cut into 2 parts. We don’t know the length at which it was cut. So, let us consider the length of the first part as ‘x’.

So, the length of the second part will be (28 - x), as 28m is the total length of wire.

It’s said that one part of the wire is converted into a square and the other part is converted into a circle.

Let us consider that the length x m is converted into a square.

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 \[x=\dfrac{28\pi }{4+\pi }\].

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?