Question

Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{a}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{and}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{be}x\mathrm{and}r,\mathrm{respectively}. \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{perimeters}\mathrm{of}\mathrm{square}\mathrm{and}\mathrm{circle}\mathrm{is}\mathrm{constant}. \\ \Rightarrow 4x+2\pi r=K\left(\mathrm{Where}K\mathrm{is}\mathrm{some}\mathrm{constant}\right) \\ \Rightarrow x=\frac{\left(K-2\pi r\right)}{4}...\left(1\right) \\ \mathrm{Now}, \\ A=x^2+\pi r^2 \\ \Rightarrow A=\frac{{\left(K-2\pi r\right)}^2}{16}+\pi r^2\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dA}{dr}=\frac{{\left(K-2\pi r\right)}^2}{16}+\pi r^2 \\ \Rightarrow \frac{dA}{dr}=\frac{2\left(K-2\pi r\right)-2\pi }{16}+2\pi r \\ \Rightarrow \frac{dA}{dr}=\frac{\left(K-2\pi r\right)-\pi }{4}+2\pi r \\ \Rightarrow \frac{\left(K-2\pi r\right)-\pi }{4}+2\pi r=0 \\ \Rightarrow \frac{\left(K-2\pi r\right)\pi }{4}=2\pi r \\ \Rightarrow K-2\pi r=8r...\left(2\right) \\ \frac{d^{2}A}{d{x}^2}=\frac{{\pi }^2}{2}+2\pi >0 \\ \mathrm{So},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{areas},\mathrm{A}\mathrm{is}\mathrm{least}\mathrm{when}K-2\pi r=8r. \\ \mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ x=\frac{\left(K-2\pi r\right)}{4} \\ \Rightarrow x=\frac{8r}{4} \\ \Rightarrow x=2r \\ \therefore \mathrm{Side}\mathrm{of}\mathrm{the}\mathrm{square}=\mathrm{Diameter}\mathrm{of}\mathrm{the}\mathrm{circle} \end{gathered}$$