Question

A wire of length $36 \,m$ 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 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)

$\frac{36\pi }{\pi + 4 }$, $\frac{144}{\pi + 4 }$

Answer 2

Let $x$ metres be the length of a side of the square and $y$ metres be the radius of the circle. Then, we have $4x + 2\pi y = 36$ $\Rightarrow 2x + \pi y = 18 \quad...\left(i\right)$ Let $A$ be the combined area of the square and the circle. Then, $A= x^{2} + \pi y^{2}\quad ...\left(ii\right)$ $\Rightarrow A = x^{2} + \pi\left(\frac{18-2x}{\pi}\right)^{2} \,$ [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, $\frac{d^{2}A}{dx^{2}} = 2 -\frac{4}{\pi }\left(-2\right) = 2 + \frac{8}{\pi}$ 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, $\left(\frac{d^{2}A}{dx^{2}}\right)_{ x = \frac{36}{\pi + 4 }} = 2 +\frac{8}{\pi} > 0$ 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 $2\pi y = 2\pi \times \frac{18}{\pi + 4 }$ $= \frac{36\pi}{\pi + 4 }m$