Question

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 lengths of the two pieces, so that the combined area of the circle and the square is minimum?

Answer 1

To solve this question, we should note the point that the sum of the perimeter of the square and circumference of the circle is equal to 28 m. We know that the perimeter of the circle with side s m is 4s m and circumference of the circle with radius r m is $=2\pi r$m. We can write the relation as $2\pi r+4s=28$ . The area of the circle $=\pi {{r}^{2}}\text{ }{{\text{m}}^{2}}$ and the area of the square $={{s}^{2}}\text{ }{{\text{m}}^{2}}$ . By summing them up, we get $A={{s}^{2}}+\pi {{r}^{2}}$ , A is the total area and we should write A in terms of any one of r and s using the equation $2\pi r+4s=28$. Derive the area equation with respect to the variable available and equate it to zero to get the value of the variable for which we get the minimum area. Use this in the perimeter equation to get both the lengths of the square piece and circular piece.

Complete step-by-step answer :

In the question, the wire AB is of length 28 m is cut into 2 pieces and made into a circle of radius r m and a square CDEF of side s m.
Circumference of the circle $=2\pi r$m
Perimeter of the circle$=4s$m
As the length of the wire is constant, sum of perimeter and circumference is equal to 28 m.
$2\pi r+4s=28$
From this we can write $r=\dfrac{28-4s}{2\pi }\to \left( 1 \right)$
We are asked to find the minimum area. Let us write the area formed by 2 figures when radius = r m and side = s m.
Area of square$={{s}^{2}}\text{ }{{\text{m}}^{2}}$
Area of circle$=\pi {{r}^{2}}\text{ }{{\text{m}}^{2}}$
Let us denote the total area as A. Total area A is given by
$A={{s}^{2}}+\pi {{r}^{2}}$
Substituting the value of r from equation-1, we get
$A={{s}^{2}}+\pi {{\left( \dfrac{28-4s}{2\pi } \right)}^{2}}$
To get the extreme value of A, we have to differentiate A with respect to s and equate it to zero to get the value of s for which we get minimum area. Differentiating gives,
$\dfrac{dA}{ds}=2s+2\pi \left( \dfrac{28-4s}{2\pi } \right)\left( \dfrac{-4}{2\pi } \right)$
Equating $\dfrac{dA}{ds}$ to zero, we get
$\begin{aligned} & \dfrac{dA}{ds}=0=2s-\dfrac{112}{2\pi }+\dfrac{16s}{2\pi } \\\ & s\left( 2+\dfrac{8}{\pi } \right)=\dfrac{56}{\pi } \\\ & s\left( \dfrac{2\pi +8}{\pi } \right)=56 \\\ & s=\dfrac{56}{8+2\pi } \\\ & s=\dfrac{28}{4+\pi } \\\ \end{aligned}$
For A to be minimum $\dfrac{{{d}^{2}}A}{d{{s}^{2}}}>0$
$\dfrac{{{d}^{2}}A}{d{{s}^{2}}}=2+\dfrac{16}{2\pi }$ which is > 0 for all values of s.
By substituting the value of s in equation-1 we get,
$\begin{aligned} & 2\pi r+4\times \dfrac{28}{4+\pi }=28 \\\ & 2\pi r=28-\dfrac{112}{4+\pi } \\\ & 2\pi r=\dfrac{112+28\pi -112}{4+\pi } \\\ \end{aligned}$

$\begin{aligned} & 2\pi r=\dfrac{28\pi }{4+\pi } \\\ & r=\dfrac{28\pi }{4+\pi }\times \dfrac{1}{2\pi } \\\ & r=\dfrac{14}{4+\pi } \\\ \end{aligned}$
The lengths of the two pieces are
Length of the square piece ${{L}_{1}}=4s=4\times \dfrac{28}{4+\pi }=\dfrac{112}{4+\pi }$
Length of the circular piece ${{L}_{2}}=2\pi r=2\pi \times \dfrac{14}{4+\pi }=\dfrac{28\pi }{4+\pi }$
$\therefore$The lengths of the square and circle are $\dfrac{112}{4+\pi }\text{ and }\dfrac{28\pi }{4+\pi }$ respectively.

Note : Students can make a mistake by taking diameter in the circumference calculation and radius in the area calculation which leads to errors. While differentiating the area function, some students forget to differentiate $\left( \dfrac{28-4s}{2\pi } \right)$ and multiply it to the term $2\pi \left( \dfrac{28-4s}{2\pi } \right)$ which leads to mistake. We have to first differentiate the whole function and then differentiate the term inside the function. The minimum area can also be calculated by substituting the calculated values of r and s in the area function.