Question

A wire length $ 2 $ unit is cut into $ 2 $ parts which are bent to form a square of side x unit and circle of radius r unit. If the sum of areas or square and circle so formed is minimum, find the relation between x and r.

Answer 1

Hint: For a given problem we first find perimeter of square and circumference of circle using given side and radius and then equating sum of both equal to length of wire to get value of variable ‘x’ in terms of r. Then using this value of ‘x’ in the equation obtained from the sum of the area of both figures to have an equation in variable ‘r’. Finally differentiating equations so obtained to get either value of ‘r’ using the concept of maxima and minima of derivatives.

Complete step-by-step answer:
Side of the square is x units. Therefore, its perimeter is given as: $ 4 \times side $
Perimeter = $ 4 \times side $
 $ \Rightarrow P = 4x $
Radius of the circle is given r units. Therefore its circumference is given as:
Circumference of circle $ = 2\pi r $
Since, it is given that wire is divided into two figures square and circle. Therefore, the sum of the perimeter of the square and circumference will be equal to the length of the wire.
 $ Lenght\,\,of\,\,wire = \,\,Perimeter\,\,of\,\,square + \,\,Circumference\,\,of\,\,the\,\,circle $
 $ \Rightarrow 2 = 4x + 2\pi r $
Taking $ 2 $ common on right hand side, we have
 $ 2 = 2\left( {2x + \pi r} \right) $
 $ \Rightarrow \dfrac{2}{2} = 2x + \pi r $
 $ \Rightarrow 1 = 2x + \pi r $
 $ \Rightarrow 2x + \pi r = 1 $
Writing ‘x’ in term of ‘r’ we have,
 $ \Rightarrow 2x = 1 - \pi r $
 $ \Rightarrow x = \dfrac{{1 - \pi r}}{2}........\left( i \right) $
Now, area of a square $ = {\left( {side} \right)^2} $
 $ \therefore $ Area of a square $ = {\left( x \right)^2} $
Area of a circle $ = \pi {r^2} $
Therefore, total area from both figures given as:
A = area of square + area of a circle
 $ \Rightarrow $ A $ = {x^2} + \pi {r^2}. $
Substituting value of ‘x’ obtained in equation (i) in above equation.
 $ A = {\left( {\dfrac{{1 - \pi r}}{2}} \right)^2} + \pi {r^2} $
Or
\[A = \dfrac{{{{\left( {1 - \pi r} \right)}^2}}}{{{{\left( 2 \right)}^2}}} + \pi {r^2}\]
\[ \Rightarrow A = \dfrac{{{{\left( {1 - \pi r} \right)}^2}}}{4} + \pi {r^2}\]
To discuss maxima or minima we find a first order derivative of ‘A’ and then equating to zero to find zero of the equation so obtained.
 $
\dfrac{{dA}}{{dr}} = \dfrac{d}{{dr}}\dfrac{{{{\left( {1 - \pi r} \right)}^2}}}{4} + \dfrac{d}{{dr}}\left( {\pi {r^2}} \right) \\
\Rightarrow \dfrac{{dA}}{{dr}} = \dfrac{1}{4}\dfrac{d}{{dr}}{\left( {1 - \pi r} \right)^2} + \dfrac{d}{{dr}}\left( {\pi {r^2}} \right) \\
\Rightarrow \dfrac{{dA}}{{dr}} = \dfrac{1}{4} \times 2(1 - \pi r)\dfrac{d}{{dr}}( - \pi r) + \pi \left( {2r} \right) \\
   \Rightarrow \dfrac{{dA}}{{dr}} = \dfrac{1}{2}\left( {1 - \pi r} \right)( - \pi ) + 2\pi r \\
 $
Equation $ \dfrac{{dA}}{{dr}} = 0 $ we have,
 $
  \dfrac{{ - 1}}{2}\left( {1 - \pi r} \right)\left( \pi \right) + 2\pi r = 0 \\
   \Rightarrow 2\pi r = \dfrac{1}{2}\left( {1 - \pi r} \right)\pi \\
   \Rightarrow 2r = \dfrac{1}{2}\left( {1 - \pi r} \right) \\
   \Rightarrow 4r = 1 - \pi r \\
   \Rightarrow 4r + \pi r = 1 \\
  or \\
  r = \dfrac{1}{{4 + \pi }} \\
  $
To discuss weather $ r = \dfrac{1}{{4 + \pi }} $ is a point of minima or maxima. We calculate the double derivative of ‘A’ if it results negative then ‘r’ is a point of maxima but if it results positive then it will be a point of minima.
 $
  \dfrac{{{d^2}A}}{{d{r^2}}} = \dfrac{d}{{dr}}\left( {\dfrac{{dA}}{{dr}}} \right) \\
\Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = \dfrac{d}{{dr}}\left( {\dfrac{1}{2}\left( {1 - \pi r} \right)( - \pi ) + 2\pi r} \right) \\
 \Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = \dfrac{{ - \pi }}{2}( - \pi ) + 2\pi \\
 \Rightarrow \dfrac{{{d^2}A}}{{d{r^2}}} = \dfrac{{{\pi ^2}}}{2} + 2\pi > 0 \\
  $
Hence from above we can say that every point or $ r = \dfrac{1}{{4 + \pi }} $ is a point of minima.
Now, substituting value of ‘r’ in equation (i) formed above to find value of ‘x’
 $
  x = \dfrac{{1 - \pi r}}{2} \\
   \Rightarrow x = \dfrac{{1 - \pi \left( {\dfrac{1}{{4 + \pi }}} \right)}}{2} \\
   \Rightarrow x = \dfrac{{1 - \dfrac{\pi }{{4 + \pi }}}}{2} \\
   \Rightarrow x = \dfrac{{\dfrac{{4 + \pi - \pi }}{{4 + \pi }}}}{2} \\
   \Rightarrow x = \dfrac{{\dfrac{4}{{4 + \pi }}}}{2} \\
   \Rightarrow x = \dfrac{2}{{4 + \pi }} \\
  $
From above we see that value of ‘x’ is double of the value of ‘r’
Therefore we can write
  $
  x = 2\left( {\dfrac{1}{{4 + r}}} \right) \\
  x = 2r \\
  $
Hence, from above we can say required relation between side of a square and radius of circle is $ x = 2r. $

Note: Maxima- minima of any function always be calculated by using derivative concept. In this we first find the first order derivative of the given function and then equate it to zero to get zeroes of the equation. After that substituting zeroes so obtained in double derivative of given function, if results so obtained is negative then point will a point of maxima and if it results positive then it will be a point of minima.