Question

A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening.

Answer 1

$$\begin{gathered} \text{Let the dimensions of the rectangular part be}\mathit{\text{x}}\text{and}\mathit{\text{y.}} \\ \mathrm{R}\text{adius of semi-circle =}\frac{x}{2} \\ \mathrm{Total}\mathrm{perimeter}=10 \\ \Rightarrow \left(x+2y\right)+\pi \left(\frac{x}{2}\right)=10 \\ \Rightarrow 2y=\left[10-x-\pi \left(\frac{x}{2}\right)\right] \\ \Rightarrow y=\frac{1}{2}\left[10-x\left(1+\frac{\pi }{2}\right)\right]...\left(1\right) \\ \mathrm{Now}, \\ \mathrm{Area},A=\frac{\pi }{2}{\left(\frac{x}{2}\right)}^2+xy \\ \Rightarrow A=\frac{\pi x^2}{8}+\frac{x}{2}\left[10-x\left(1+\frac{\pi }{2}\right)\right]\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow A=\frac{\pi x^2}{8}+\frac{10x}{2}-\frac{x^2}{2}\left(1+\frac{\pi }{2}\right) \\ \Rightarrow \frac{dA}{dx}=\frac{\pi x}{4}+\frac{10}{2}-\frac{2x}{2}\left(1+\frac{\pi }{2}\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}A,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dA}{dx}=0 \\ \Rightarrow \frac{\pi x}{4}+\frac{10}{2}-\frac{2x}{2}\left(1+\frac{\pi }{2}\right)=0 \\ \Rightarrow x\left[\frac{\pi }{4}-1-\frac{\pi }{2}\right]=-5 \\ \Rightarrow x=\frac{-5}{\left({\displaystyle \frac{-4-\pi }{4}}\right)} \\ \Rightarrow x=\frac{20}{\left(\pi +4\right)} \\ \text{Substituting the value of}\mathit{\text{x}}\text{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ y=\frac{1}{2}\left[10-\left(\frac{20}{\pi +4}\right)\left(1+\frac{\pi }{2}\right)\right] \\ \Rightarrow y=5-\frac{10\left(\pi +2\right)}{2\left(\pi +4\right)} \\ \Rightarrow y=\frac{5\pi +20-5\pi -10}{\left(\pi +4\right)} \\ \Rightarrow y=\frac{10}{\left(\pi +4\right)} \\ \frac{d^{2}A}{d{x}^2}=\frac{\pi }{4}-\frac{\pi }{2}-1 \\ \Rightarrow \frac{d^{2}A}{d{x}^2}=\frac{\pi -2\pi -4}{4} \\ \Rightarrow \frac{d^{2}A}{d{x}^2}=\frac{-\pi -4}{4}<0 \\ \mathrm{Thus},\mathrm{the}\mathrm{a}\text{rea is maximum when}\mathit{\text{x=}}\frac{20}{\pi +4}\text{and}\mathit{\text{y}}\text{=}\frac{10}{\pi +4}\text{.} \\ \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{dimensions}\mathrm{are}\mathrm{given}\mathrm{below}: \\ \mathrm{Length}=\frac{20}{\mathrm{\pi }+4}\mathrm{m} \\ \mathrm{B}\text{readth=}\frac{10}{\mathrm{\pi }+4}\text{m} \end{gathered}$$