Question

A wire of length l is cut into pieces. One of the pieces is made into a circle and the other into a square then sum of the area of the circle and the square is least if the radius of the circle is $\dfrac{1}{{6 + 2\pi }}$ units.
A.True
B.False

Answer 1

Hint : In this question, we are given a straight wire of length “l” and then this wire is cut into two parts to make a circle and square with each part. We will let the length of one of the parts be “x” and then the length of the part will be “l-x”. The length of the part of the wire will be equal to the perimeter of the shape formed by it. We can solve this question easily by using the above information.

Complete step by step solution:
Let the length of the part of the wire made into a circle be “x”, so the length of the wire made into a square is “l-x”.
So, we have –
$2\pi r = x$ and $4s = l - x$
$\Rightarrow r = \dfrac{x}{{2\pi }}$ and $s = \dfrac{{l - x}}{4}$
Area of circle is ${A_1} = \pi {(\dfrac{x}{{2\pi }})^2} = \dfrac{{{x^2}}}{{4\pi }}$ and Area of square is ${A_2} = {(\dfrac{{l - x}}{4})^2} = \dfrac{{{l^2} + {x^2} - 2lx}}{{16}}$
Sum of the area of the circle and the square is $A = \dfrac{{{x^2}}}{{4\pi }} + \dfrac{{{l^2} + {x^2} - 2lx}}{{16}}$
Differentiating both sides with respect to x, we get –
$\dfrac{{dA}}{{dx}} = \dfrac{1}{{4\pi }}(2x) + \dfrac{1}{{16}}(2x - 2l) \\\ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{x}{{2\pi }} + \dfrac{{x - l}}{8} \;$
To find the value of x at which the sum of areas is minimum, we put the above equation equal to 0 –
$\dfrac{x}{{2\pi }} + \dfrac{{x - l}}{8} = 0 \\\ \Rightarrow \dfrac{{4x + x\pi - l\pi }}{{8\pi }} = 0 \\\ \Rightarrow (4 + \pi )x = l\pi \\\ \Rightarrow x = \dfrac{{l\pi }}{{4 + \pi }} \;$
So, the value of radius when the area is minimum is –
$r = \dfrac{x}{{2\pi }} = \dfrac{{\dfrac{{l\pi }}{{4 + \pi }}}}{{2\pi }} = \dfrac{l}{{8 + 2\pi }}$
So the answer is false.
Hence option (B) is the correct answer.
So, the correct answer is “Option B”.

Note : Perimeter of a shape is defined as the length of the boundary of the shape. As the shape is formed by a piece of wire, the length of the boundary will be equal to the length of that piece of wire. To verify that the area is minimum for $x = \dfrac{{l\pi }}{{4 + \pi }}$ , we can do a second derivative test of Area.
$\dfrac{{{d^2}A}}{{d{x^2}}} = \dfrac{1}{{2\pi }} + \dfrac{1}{8}$
Clearly $\dfrac{{{d^2}A}}{{d{x^2}}} > 0$ , so the area is minimum at $x = \dfrac{{l\pi }}{{4 + \pi }}$ .