Question

(A wire of length 28 cm is to be cut into two picces. One of the wires
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 ?
CBSE 1996 give me
answer ​

Answer 1

Given:

A wire of length 28 cm is cut down into 2 pieces

One of the wires is to be made into a square and the other into a circle

To Find:

The length of the two pieces in the manner that the combined area of the square and the circle is minimum.

Solution:

We know that,

• Area of circle is given by

$\pink{\boxed{\bf{Area\:of\:Circle=\pi r^{2}}}}$

where r is the radius of circle

• Area of square is given by

$\purple{\boxed{\bf{Area\:of\:Square=(a)^{2}}}}$

where a is length of the side of square

$\rule{190}{1}$

Let the length of the piece which is drawn in circle be x cm

Then the length of other piece which is drawn in square will be (28-x) cm

Now,

Let the radius of circle be r and length of side of square be a

So, according to question

$\rightarrow\rm{Perimeter\:of\:cicle=Length\:of\:first\:piece}$

$\rightarrow\rm{2\pi r=x}$

$\rightarrow\rm{r=\dfrac{x}{2\pi}}$

Also,

$\rightarrow\rm{Perimeter\:of\:square=Length\:of\:second\:piece}$

$\rightarrow\rm{4a=28-x}$

$\rightarrow\rm{a=\dfrac{28-x}{4}}$

$\rule{190}{1}$

Now,

Let the area of the circle be A₁

So,

$\longrightarrow\rm{A_{1}=\pi r^{2}}$

$\longrightarrow\rm{A_{1}=\pi\bigg(\dfrac{x}{2\pi}\bigg)^{2}}$

$\longrightarrow\rm{A_{1}=\dfrac{\pi x^{2}}{4\pi^{2}}}$

$\longrightarrow\rm{A_{1}=\dfrac{x^{2}}{4\pi}}$

Also, let the area of the square be A₂

So,

$\longrightarrow\rm{A_{2}=(a)^{2}}$

$\longrightarrow\rm{A_{2}=\bigg(\dfrac{28-x}{4}\bigg)^{2}}$

$\longrightarrow\rm{A_{2}=\dfrac{(28-x)^{2}}{16}}$

$\rule{190}{1}$

Now, let the combined area of circle and square be A

So,

$\longrightarrow\rm{A=A_{1}+A_{2}}$

$\longrightarrow\rm{A=\dfrac{x^{2}}{4\pi}+\dfrac{(28-x)^{2}}{16}}$

On differentiating A with respect to x, we get

$\longrightarrow\rm{\dfrac{dA}{dx}=\dfrac{\cancel{2}x}{\cancel{4}\pi}+\dfrac{\cancel{-2}(28-x)}{\cancel{16}}}$

$\longrightarrow\rm{\dfrac{dA}{dx}=\dfrac{x}{2\pi}+\dfrac{-(28-x)}{8}}$

$\longrightarrow\rm{\dfrac{dA}{dx}=\dfrac{x}{2\pi}-\dfrac{28-x}{8}}$

To find the minimum value of A, we have to let dA/dx=0

So, let

$\longrightarrow\rm{\dfrac{dA}{dx}=0}$

$\longrightarrow\rm{\dfrac{x}{2\pi}-\dfrac{28-x}{8}=0}$

$\longrightarrow\rm{\dfrac{x}{\cancel{2}\pi}=\dfrac{28-x}{\cancel{8}}}$

$\longrightarrow\rm{\dfrac{x}{\pi}=\dfrac{28-x}{4}}$

$\longrightarrow\rm{4x=\pi(28-x)}$

$\longrightarrow\rm{4x=28\pi-\pi x}$

$\longrightarrow\rm{\pi x+4x=28\pi}$

$\longrightarrow\rm{x(\pi+4)=28\pi}$

$\longrightarrow\rm{x=\dfrac{28\pi}{\pi+4}}$

$\rule{190}{1}$

Now, again differentiating A with respect to x, we get

$\longrightarrow\rm{\dfrac{d^{2}A}{dx^{2}}=\dfrac{1}{2\pi}-\dfrac{(-1)}{8}}$

$\longrightarrow\rm{\dfrac{d^{2}A}{dx^{2}}=\dfrac{1}{2\pi}+\dfrac{1}{8}=+ve}$

Since, double differentiation of A is positive, so A is minimum for the obtained value of x

$\rule{190}{1}$

So, the length of two pieces are

$\rm\green{Length\:of\:first\:piece=x=\dfrac{28\pi}{\pi+4}}$

$\rm{Length\:of\:second\:piece=28-x}$

$\rm{Length\:of\:second\:piece=28-\dfrac{28\pi}{\pi+4}}$

$\rm{Length\:of\:second\:piece=\dfrac{28\pi+112-28\pi}{\pi+4}}$

$\rm\green{Length\:of\:second\:piece=\dfrac{112}{\pi+4}}$

Answer 2

Let one side of the square thus formed be $\displaystyle\sf{a}$ and the radius of the circle formed be $\displaystyle\sf{r.}$

Here the sum of perimeters of square and circle should be equal to length of wire, 28 cm.

$\displaystyle\longrightarrow\sf{4a+2\pi r=28}$

$\displaystyle\longrightarrow\sf{2\pi r=28-4a}$

$\displaystyle\longrightarrow\sf{\pi r=14-2a}$

$\displaystyle\longrightarrow\sf{(\pi r)^2=(14-2a)^2}$

$\displaystyle\longrightarrow\sf{\pi^2r^2=(14-2a)^2}$

$\displaystyle\longrightarrow\sf{\pi r^2=\dfrac {(14-2a)^2}{\pi}\quad\quad\dots (1)}$

The combined area of square and circle is the sum of their areas.

$\displaystyle\longrightarrow\sf{A=a^2+\pi r^2}$

From (1),

$\displaystyle\longrightarrow\sf{A=a^2+\dfrac {(14-2a)^2}{\pi}}$

For combined area to have minimum value, its derivative with respect to $\displaystyle\sf{a}$ should be taken as zero.

$\displaystyle\longrightarrow\sf{\dfrac {dA}{da}=0}$

$\displaystyle\longrightarrow\sf{\dfrac {d}{da}\,\left [a^2+\dfrac {(14-2a)^2}{\pi}\right]=0}$

$\displaystyle\longrightarrow\sf{2a+\dfrac {2(14-2a)(-2)}{\pi}=0}$

$\displaystyle\longrightarrow\sf{2a-\dfrac {4(14-2a)}{\pi}=0}$

$\displaystyle\longrightarrow\sf{2a\pi-4(14-2a)=0}$

$\displaystyle\longrightarrow\sf{a\pi-2(14-2a)=0}$

$\displaystyle\longrightarrow\sf{a(\pi+4)-28=0}$

$\displaystyle\longrightarrow\sf{a=\dfrac {28}{\pi+4}}$

$\displaystyle\longrightarrow\underline {\underline{\sf{4a=\dfrac {112}{\pi+4}\ cm}}}$

Taking $\displaystyle\sf{\pi=3.14,}$

$\displaystyle\longrightarrow\underline {\underline{\sf{4a=15.68\ cm}}}$

And,

$\displaystyle\longrightarrow\sf{2\pi r=28-4a}$

$\displaystyle\longrightarrow\sf{2\pi r=28-\dfrac{112}{\pi+4}}$

$\displaystyle\longrightarrow\underline {\underline {\sf{2\pi r=\dfrac {28\pi}{\pi+4}\ cm}}}$

$\displaystyle\longrightarrow\underline {\underline {\sf{2\pi r=12.32\ cm}}}$

Hence the wire should be cut as $\displaystyle\bf{15.68\ cm}$ for making square and $\displaystyle\bf{12.32\ cm}$ for making square.