Question

A wire of length 28 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 circle and the square is minimum?

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{the}\mathrm{wire},\mathrm{which}\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{made}\mathrm{into}\mathrm{a}\mathrm{square}\mathrm{and}\mathrm{a}\mathrm{circle},\mathrm{is}\mathrm{cut}\mathrm{into}\mathrm{two}\mathrm{pieces}\mathrm{of}\mathrm{length}x\mathrm{m}\mathrm{and}y\mathrm{m},\mathrm{respectively}.\mathrm{Then}, \\ x+y=28...\left(1\right) \\ \mathrm{Perimeter}\mathrm{of}\mathrm{square},4\left(\mathrm{side}\right)=x \\ \Rightarrow \mathrm{Side}=\frac{x}{4} \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{square}={\left(\frac{x}{4}\right)}^2=\frac{x^2}{16} \\ \mathrm{Circumference}\mathrm{of}\mathrm{circle},2\mathrm{\pi }r\mathit{=}y \\ \Rightarrow r=\frac{y}{2\mathrm{\pi }} \\ \mathrm{Area\; of\; circle}=\mathrm{\pi }{r}^2=\mathrm{\pi }{\left(\frac{y}{2\mathrm{\pi }}\right)}^2=\frac{y^2}{4\mathrm{\pi }} \\ \mathrm{Now}, \\ z=\mathrm{Area}\mathrm{of}\mathrm{square}+\mathrm{Area\; of\; circle} \\ \Rightarrow z=\frac{x^2}{16}+\frac{y^2}{4\mathrm{\pi }} \\ \Rightarrow z=\frac{x^2}{16}+\frac{{\left(28-x\right)}^2}{4\mathrm{\pi }} \\ \Rightarrow \frac{dz}{dx}=\frac{2x}{16}-\frac{2\left(28-x\right)}{4\mathrm{\pi }} \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dz}{dx}=0 \\ \Rightarrow \frac{2x}{16}-\frac{2\left(28-x\right)}{4\mathrm{\pi }}=0\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{x}{4}=\frac{\left(28-x\right)}{\mathrm{\pi }} \\ \Rightarrow \frac{x\mathrm{\pi }}{4}+x=28 \\ \Rightarrow x\left(\frac{\mathrm{\pi }}{4}+1\right)=28 \\ \Rightarrow x=\frac{28}{\left({\displaystyle \frac{\mathrm{\pi }}{4}+1}\right)} \\ \Rightarrow x=\frac{112}{\mathrm{\pi }+4} \\ \Rightarrow y=28-\frac{112}{\mathrm{\pi }+4}\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow y=\frac{28\pi }{\mathrm{\pi }+4} \\ \frac{d^{2}z}{d{x}^2}=\frac{1}{8}+\frac{1}{2\mathrm{\pi }}>0 \\ \mathrm{Thus},z\mathrm{is}\mathrm{minimum}\mathrm{when}x=\frac{112}{\mathrm{\pi }+4}\mathrm{and}y=\frac{28\mathrm{\pi }}{\mathrm{\pi }+4}. \\ \mathrm{Hence},\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{pieces}\mathrm{of}\mathrm{wire}\mathrm{are}\frac{112}{\mathrm{\pi }+4}\mathrm{m}\mathrm{and}\frac{28\mathrm{\pi }}{\mathrm{\pi }+4}\mathrm{m}\mathrm{respectively}. \end{gathered}$$