Question

A wire of length $36m$ 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 square and the circle is minimum?

• A. $\frac{36\pi }{\pi +4}$, $\frac{36\pi }{\pi +4}$
• B. $\frac{36\pi }{\pi +4}$, $\frac{144}{\pi +4}$
• C. $\frac{36\pi }{\pi +4}$, $\frac{144\pi }{\pi +4}$
• D. $\frac{36}{\pi +4}$, $\frac{125}{\pi +4}$

Answer 1

Answer: (B)

Let $x$ metres be the length of a side of the square and $y$ metres be the radius of the circle. Then, we have
$4x+2\pi y=36$
$\Rightarrow 2x+\pi y=18...\left(i\right)$
Let $A$ be the combined area of the square and the circle. Then,
$A=x^2+\pi y^2...\left(ii\right)$
$\Rightarrow A=x^2+\pi {\left(\frac{18-2x}{\pi }\right)}^2$ [Using $\left(i\right)$]
$\Rightarrow A=x^2+\frac{1}{\pi }{\left(18-2x\right)}^2$
$\Rightarrow \frac{dA}{dx}=2x+\frac{2}{\pi }\left(18-2x\right)\left(-2\right)=2x-\frac{4}{\pi }\left(18-2x\right)$
and, $\frac{d^{2}A}{d{x}^2}=2-\frac{4}{\pi }\left(-2\right)=2+\frac{8}{\pi }$
For maximum or minimum $A$, we must have
$\frac{dA}{dx}=0$
$\Rightarrow 2x+\frac{4}{\pi }\left(18-2x\right)=0$
$\Rightarrow x=\frac{36}{\pi +4}$
Clearly, ${\left(\frac{d^{2}A}{d{x}^2}\right)}_{x=\frac{36}{\pi +4}}=2+\frac{8}{\pi }>0$
Thus, $A$ is minimum when $x=\frac{36}{\pi +4}$
Putting $x=\frac{36}{\pi +4}$ in $\left(i\right)$, we obtain $y=\frac{18}{\pi +4}$
So, lengths of the two pieces of wire are
$4x=4\times \frac{36}{\pi +4}=\frac{144}{\pi +4}m$
and $2\pi y=2\pi \times \frac{18}{\pi +4}$
$=\frac{36\pi }{\pi +4}m$