Question

A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{dimensions}\mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{be}x\mathrm{and}y. \\ \mathrm{Perimeter}\mathrm{of}\mathrm{the}\mathrm{window}=x+y+x+x+y=12 \\ \Rightarrow 3x+2y=12 \\ \Rightarrow y=\frac{12-3x}{2}...\left(1\right) \\ \text{Area of the window =}\mathit{\text{xy}}\text{+}\frac{\sqrt{3}}{4}{\text{x}}^2 \\ \Rightarrow A=x\left(\frac{12-3x}{2}\right)+\frac{\sqrt{3}}{4}{x}^2 \\ \Rightarrow A=6x-\frac{3x^2}{2}+\frac{\sqrt{3}}{4}{x}^2 \\ \Rightarrow \frac{dA}{dx}=6-\frac{6x}{2}+\frac{2\sqrt{3}}{4}x \\ \Rightarrow \frac{dA}{dx}=6-3x+\frac{\sqrt{3}}{2}x \\ \Rightarrow \frac{dA}{dx}=6-x\left(3-\frac{\sqrt{3}}{2}\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{minimum}\mathrm{values}\mathrm{of}\mathrm{A},\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dA}{dx}=0 \\ \Rightarrow 6=x\left(3-\frac{\sqrt{3}}{2}\right) \\ \Rightarrow x=\frac{12}{6-\sqrt{3}} \\ \text{Substituting the value of}\mathit{\text{x}}\text{in}\mathrm{eq}.\left(\text{1}\right),\mathrm{we}\mathrm{get} \\ y=\frac{12-3\left({\displaystyle \frac{12}{6-\sqrt{3}}}\right)}{2} \\ \Rightarrow y=\frac{18-6\sqrt{3}}{6-\sqrt{3}} \\ \mathrm{Now}, \\ \frac{d^{2}A}{d{x}^2}=-3+\frac{\sqrt{3}}{2}<0 \\ \text{Thus, the area is maximum when}\mathit{\text{x}}\text{=}\frac{12}{6-\sqrt{3}}\text{and}\mathit{\text{y}}\text{=}\frac{18-6\sqrt{3}}{6-\sqrt{3}}\text{.} \end{gathered}$$

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.