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

Let the dimensions of the rectangular part be x and y.
Perimeter of the window = x + y + x + x + y = 12
⇒ 3x + 2y = 12
$\text{y}=\frac{12-3\text{x}}{2}....(\text{i})$
Area of the window $=\text{xy}+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\text{A}=\text{x}\Big(\frac{12-3\text{x}}{2}\Big)+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\text{A}=6\text{x}-\frac{3\text{x}^{2}}{2}+\frac{\sqrt{3}}{4}\text{x}^{2}$
$\Rightarrow\frac{\text{dA}}{\text{dx}}=6\text{x}-\frac{6\text{x}}{2}+\frac{2\sqrt{3}}{4}\text{x}$
For maximum or minimum values of A, We must have $\frac{\text{dA}}{\text{dx}}=0$
$\Rightarrow 6=\text{x}\Big(3-\frac{\sqrt{3}}{2}\Big) $
$\Rightarrow \text{x}=\frac{12}{6\sqrt{3}} $
Substituting the values of x in eq.(i), We get
$\Rightarrow \text{y}=\frac{12-3\Big(\frac{12}{6-\sqrt{3}}\Big)}{2} $
$\Rightarrow \text{y}=\frac{18-6\sqrt{3}}{6-\sqrt{3}}$.

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