Question

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side $=x$ units and a circle of radius $=r$ units. If the sum of the areas of the square and the circle so formed is minimum, then: \[\]
A. $2c=r$\[\]
B. $2x=\left( \pi +4 \right)r$\[\]
C. $\left( 4-\pi \right)x=\pi r$\[\]
D. $x=2r$\[\]

Answer 1

Hint: We see that the sum of perimeter of the square ${{P}_{s}}$ and the perimeter of the circle ${{P}_{c}}$ is the length of the wire of 2 units. We use ${{P}_{s}}+{{P}_{c}}=2$ to get an expression of $r$ in terms of $x$. We put the expression in total area $A\left( x \right)={{A}_{s}}+{{A}_{c}}$ and find the critical point say $x=c$ from the equation ${{A}^{'}}\left( x \right)=0$. We use the first derivative test to prove the area will be minimum at $x=c$ and the find $r$ by putting $x=c$ in the expression of $r$. We compare $x$ and $r$ to choose the correct option.

Complete step-by-step solution:
We know from the first derivative test that the critical points of a function are the points where the first-order derivative of the function is zero or is not defined. Mathematically, $x=c$ is a critical point if $f'(c)=0$ or $f'(c)$ does not exist. The minima or maxima occur only at critical points. The first derivative test tells us that $f\left( x \right)$ has a local minima at $x=c$ when ${{f}^{'}}\left( x \right)$ changes sign from negative to positive as we increase though the point $x=c$.
We are given the question that a wire of length 2 units is cut into two parts which are bent respectively to form a square of side $=x$ units and a circle of radius $=r$ units. So the sum of the perimeter of the square ${{P}_{s}}$ and the perimeter of the circle ${{P}_{c}}$ is the length of the wire. So we have
\[\begin{align}
  & {{P}_{s}}+{{P}_{c}}=2 \\
 & \Rightarrow 4x+2\pi r=2 \\
 & \Rightarrow 2\pi r=2-4x \\
 & \Rightarrow r=\dfrac{1-2x}{\pi }....\left( 1 \right) \\
\end{align}\]

So the sum of the area of square and circle in square units is
\[\begin{align}
  & A={{A}_{s}}+{{A}_{c}} \\
 & \Rightarrow A={{x}^{2}}+\pi {{r}^{2}} \\
 & \Rightarrow A={{x}^{2}}+\pi \times {\left( \dfrac{1-2x}{{\pi }} \right)}^{2} \\
 & \Rightarrow A\left( x \right)={{x}^{2}}+\dfrac{{{\left( 1-2x \right)}^{2}}}{\pi } \\
\end{align}\]
We are asked to find the relation between $x$ and $r$ when the total area $A\left( x \right)$ is minimum. Let us find the critical points by equating the differentiation of $A\left( x \right)$ that is ${{A}^{'}}\left( x \right)$ to 0. We have,
\[\begin{align}
  & \therefore {{A}^{'}}\left( x \right)=\dfrac{d}{dx}A\left( x \right)=0 \\
 & \Rightarrow \dfrac{d}{dx}\left[ {{x}^{2}}+\dfrac{{{\left( 1-2x \right)}^{2}}}{\pi } \right]=0 \\
 & \Rightarrow 2x+\dfrac{2}{\pi }\left( 1-2x \right)\left( -2 \right)=0 \\
 & \Rightarrow x+\dfrac{-2+4x}{\pi }=0 \\
\end{align}\]
We simplify the above step by multiplying $\pi $ in all the terms and proceed to have the critical point as ,
\[\begin{align}
  & \Rightarrow x\pi -2+4x=0 \\
 & \Rightarrow x\left( \pi +4 \right)=2 \\
 & \Rightarrow x=\dfrac{2}{\pi +4}....\left( 2 \right) \\
\end{align}\]
Let us put two points before and after critical point $x=\dfrac{2}{\pi +4}$ as $x=0< \dfrac{2}{\pi +4}$ and $x=1> \dfrac{2}{\pi +4}$ in ${{A}^{'}}\left( x \right)$ to check the first derivative test. We have
\[\begin{align}
  & {{\left. {{A}^{'}}\left( x \right) \right|}_{x=0}}=2\times 0+\dfrac{2}{\pi }\left( 1-2\times 0 \right)\left( -2 \right)< 0 \\
 & {{\left. {{A}^{'}}\left( x \right) \right|}_{x=1}}=2\times 1+\dfrac{2}{\pi }\left( 1-2\times 1 \right)\left( -2 \right)=2+\dfrac{4}{\pi }> 0 \\
\end{align}\]
So the first derivative ${{A}^{'}}\left( x \right)$ changes sign from negative to positive through $x=\dfrac{2}{\pi +4}$. So there is a local minimum at $x=\dfrac{2}{\pi +4}$. We put $x=\dfrac{2}{\pi +4}$ in equation (1) and have,
\[r=\dfrac{1-2x}{\pi }=\dfrac{1-2\left( \dfrac{2}{\pi +4} \right)}{\pi }=\dfrac{\pi +4-4}{\left( \pi +4 \right)\pi }=\dfrac{1}{\pi +4}....\left( 3 \right)\]
We compare right hand side of equations (2) and (3) and conclude that
\[x=2r\]
So the correct option is D.

Note: We note that while taking two points before and after the critical point $x=c$ we must take them as close as possible because there can be more than one minimum for the function. The area function $A\left( x \right)={{x}^{2}}+\dfrac{{{\left( 1-2x \right)}^{2}}}{\pi }$ here is a upward parabolic function and like all upward parabolic function it has one critical point with local minima. We also note that the question assumes that the thickness of the wire to be approximately zero.