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)

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$