Question

A wire of length 50m is cut into two pieces. One piece of the wire is bent into the shape of the square and another in the shape of the circle. What should be the length of each piece so that the combined area of the two is minimum.

Answer 1

Hint: First, we should know the formula for the perimeter of square and circle and also the area of square and circle. Then, we have to consider the length of the one piece for the shape of the square is x m and the length of the other piece for the shape of the circle is (50-x)m. Then, we calculate the combined area of both the figures. Then, to get the minimum area, we need to differentiate the above expression with respect to x. Then, to get the minimum, we will use the condition as $\dfrac{dA}{dx}=0$.
After that we will get the value of both length of square wire and circle wire.

Complete step-by-step answer:
In this question, we are supposed to find the value of the length of each piece so that the combined area of the two is minimum from the two figures that are square and circle.
So, we are given that the length of the wire is 50m.
Now, let the length of the one piece for the shape of the square is x m.
Then, according to the given condition in the question, the length of the other piece for the shape of the circle is (50-x)m.
So, we know the formula for the perimeter of the square with side a is given by:
Perimeter of the square=4a
Now, we know from the above condition that the total length of all four sides of the square is x.
Then, by using this condition we get:
$\begin{align}
  & 4a=x \\
 & \Rightarrow a=\dfrac{x}{4} \\
\end{align}$

Then, similarly we know the formula for the circumference of the circle with radius r is given by:
Circumference of a circle=$2\pi r$
Now, we know from the above condition that the total length of all circles is (50-x).
Then, by using this condition we get:
$\begin{align}
  & 2\pi r=\left( 50-x \right) \\
 & \Rightarrow r=\dfrac{50-x}{2\pi } \\
\end{align}$

Now by using the formula for the area of the circle and square is:
Area of square=${{a}^{2}}$
Area of the circle=$\pi {{r}^{2}}$
Now, the combined area(A) after substituting the value of a as $\dfrac{x}{4}$ and r as $\dfrac{50-x}{2\pi }$:
$\begin{align}
  & A={{a}^{2}}+\pi {{r}^{2}} \\
 & \Rightarrow A=\dfrac{{{x}^{2}}}{16}+\pi {{\left( \dfrac{50-x}{2\pi } \right)}^{2}} \\
 & \Rightarrow A=\dfrac{{{x}^{2}}}{16}+\pi \dfrac{{{\left( 50-x \right)}^{2}}}{4{{\pi }^{2}}} \\
 & \Rightarrow A=\dfrac{{{x}^{2}}}{16}+\dfrac{{{\left( 50-x \right)}^{2}}}{4\pi } \\
\end{align}$
Now to get the minimum area, we need to differentiate the above expression with respect to x as:
$\begin{align}
  & \dfrac{dA}{dx}=\dfrac{2x}{16}+\dfrac{2\left( 50-x \right)\left( -1 \right)}{4\pi } \\
 & \Rightarrow \dfrac{dA}{dx}=\dfrac{x}{8}+\dfrac{\left( x-50 \right)}{2\pi } \\
 & \Rightarrow \dfrac{dA}{dx}=\dfrac{\pi x+4x-200}{8\pi } \\
 & \Rightarrow \dfrac{dA}{dx}=\dfrac{x\left( 4+\pi \right)-200}{8\pi } \\
\end{align}$
Now, to get the minimum, we will use the condition as $\dfrac{dA}{dx}=0$.
So, the above expression is equated to zero as:
$\begin{align}
  & \dfrac{x\left( 4+\pi \right)-200}{8\pi }=0 \\
 & \Rightarrow x\left( 4+\pi \right)-200=0 \\
 & \Rightarrow x\left( 4+\pi \right)=200 \\
 & \Rightarrow x=\dfrac{200}{\left( 4+\pi \right)} \\
\end{align}$
So, we get the value of x where the minimum value is calculated as $\dfrac{200}{\left( 4+\pi \right)}$.
So, the length of the square wire is $\dfrac{200}{\left( 4+\pi \right)}$m.
Then, the length of the circle wire is 50-x and substitute the x value in it, we get:
$\begin{align}
  & 50-\dfrac{200}{\left( 4+\pi \right)}=\dfrac{200+50\pi -200}{\left( 4+\pi \right)} \\
 & \Rightarrow \dfrac{50\pi }{\left( 4+\pi \right)} \\
\end{align}$
Hence, the length of square is $\dfrac{200}{\left( 4+\pi \right)}$m and circle wire is $\dfrac{50\pi }{\left( 4+\pi \right)}$m to get the minimum area.

Note: For solving these types of questions we must be aware of some of the basic formulas of the basic figures like square, rectangle, and circle which are frequently used in the questions. Some of them are as follows:
 Perimeter of the square=4a
Circumference of a circle=$2\pi r$
Area of square=${{a}^{2}}$
Area of the circle=$\pi {{r}^{2}}$