#### 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.

#### Solution

\[\text { Let the dimensions of the rectangle be x and y } . \]

\[\text { Perimeter of the window = x + y + x + x + y = 12 }\]

\[ \Rightarrow 3x + 2y = 12\]

\[ \Rightarrow y = \frac{12 - 3x}{2} ...........\left( 1 \right)\]

\[\text { Area of the window } =xy+\frac{\sqrt{3}}{4} x^2 \]

\[ \Rightarrow A = x\left( \frac{12 - 3x}{2} \right) + \frac{\sqrt{3}}{4} x^2 \]

\[ \Rightarrow A = 6x - \frac{3 x^2}{2} + \frac{\sqrt{3}}{4} x^2 \]

\[ \Rightarrow \frac{dA}{dx} = 6 - \frac{6x}{2} + \frac{2\sqrt{3}}{4}x\]

\[ \Rightarrow \frac{dA}{dx} = 6 - 3x + \frac{\sqrt{3}}{2}x\]

\[ \Rightarrow \frac{dA}{dx} = 6 - x\left( 3 - \frac{\sqrt{3}}{2} \right)\]

\[\text { For maximum or a minimum values of A, we must have }\]

\[\frac{dA}{dx} = 0\]

\[ \Rightarrow 6 = x\left( 3 - \frac{\sqrt{3}}{2} \right)\]

\[ \Rightarrow x = \frac{12}{6 - \sqrt{3}}\]

\[\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }\]

\[y = \frac{12 - 3\left( \frac{12}{6 - \sqrt{3}} \right)}{2}\]

\[ \Rightarrow y = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}\]

\[\text { Now, }\]

\[\frac{d^2 A}{d x^2} = - 3 + \frac{\sqrt{3}}{2} < 0\]

\[\text { Thus, the area is maximum when }x=\frac{12}{6 - \sqrt{3}}\text { and }y=\frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}.\]

#### Notes

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