Question

A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle 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{triangle},\mathrm{is}\mathrm{cut}\mathrm{into}\mathrm{two}\mathrm{pieces}\mathrm{of}\mathrm{length}x\mathrm{and}y,\mathrm{respectively}.\mathrm{Then}, \\ x+y=20...\left(1\right) \\ \mathrm{Perimeter}\mathrm{of}\mathrm{square},4\left(\mathrm{Side}\right)=x \\ \Rightarrow \mathrm{Side}=\frac{x}{\mathit{4}} \\ \mathrm{Area}\mathrm{of}\mathrm{square}={\left(\frac{x}{\mathit{4}}\right)}^2=\frac{{\mathrm{x}}^2}{16} \\ \mathrm{Perimeter}\mathrm{of}\mathrm{triangle},3\left(\mathrm{Side}\right)=y \\ \Rightarrow \mathrm{Side}=\frac{y}{3} \\ \mathrm{Area\; of}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2=\frac{\sqrt{3}}{4}\times {\left(\frac{y}{3}\right)}^2=\frac{\sqrt{3}{y}^2}{36} \\ \mathrm{Now}, \\ z=\mathrm{Area}\mathrm{of}\mathrm{square}+\mathrm{Area\; of}\mathrm{triangle} \\ \Rightarrow z=\frac{x^2}{16}+\frac{\sqrt{3}{y}^2}{36} \\ \Rightarrow z=\frac{x^2}{16}+\frac{\sqrt{3}{\left(20-x\right)}^2}{36}\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dz}{dx}=\frac{2x}{16}-\frac{2\sqrt{3}\left(20-x\right)}{36} \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}\mathrm{z},\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dz}{dx}=0 \\ \Rightarrow \frac{2x}{16}-\frac{\sqrt{3}\left(20-x\right)}{18}=0 \\ \Rightarrow \frac{9x}{4}=\sqrt{3}\left(20-x\right) \\ \Rightarrow \frac{9x}{4}+x\sqrt{3}=20\sqrt{3} \\ \Rightarrow x\left(\frac{9}{4}+\sqrt{3}\right)=20\sqrt{3} \\ \Rightarrow x=\frac{20\sqrt{3}}{\left({\displaystyle \frac{9}{4}}+\sqrt{3}\right)} \\ \Rightarrow x=\frac{80\sqrt{3}}{\left(9+4\sqrt{3}\right)} \\ \Rightarrow y=20-\frac{80\sqrt{3}}{9+4\sqrt{3}}\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow y=\frac{180}{9+4\sqrt{3}} \\ \frac{d^{2}z}{d{x}^2}=\frac{1}{8}+\frac{\sqrt{3}}{18}>0 \\ \mathrm{Thus},z\mathrm{is}\mathrm{minimum}\mathrm{when}x=\frac{80\sqrt{3}}{\left(9+4\sqrt{3}\right)}\mathrm{and}y=\frac{180}{9+4\sqrt{3}}. \\ \mathrm{Hence},\mathrm{the}\mathrm{wire}\mathrm{of}\mathrm{length}20\mathrm{cm}\mathrm{should}\mathrm{be}\mathrm{cut}\mathrm{into}\mathrm{two}\mathrm{pieces}\mathrm{of}\mathrm{lengths}\frac{80\sqrt{3}}{\left(9+4\sqrt{3}\right)}\mathrm{m}\mathrm{and}\frac{180}{9+4\sqrt{3}}\mathrm{m}. \\ \mathrm{Disclaimer}:\mathrm{The}\mathrm{solution}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}\mathrm{is}\mathrm{incorrrect}.\mathrm{The}\mathrm{solution}\mathrm{here}\mathrm{is}\mathrm{created}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}. \end{gathered}$$