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

Answer


let sides of rectangle be x and y and the sides of equilateral triangle be x
$\therefore$3x + 2y = 12 $\Rightarrow$ $\text{y}=\frac{\text{12 - 3x}}{2}$
Area = xy + $\sqrt{3}\frac{\text{x}^{2}}{\text{4}}$
$=\text{x}\frac{\text{(12 - 3x)}}{2}+\sqrt{3}\frac{\text{x}^{2}}{4}$
$\text{A}=\frac{1}{4}[24\text{x - 6x}^{2}+\sqrt{3}\text{ x}^{2}]$
$\frac{\text{dA}}{\text{dx}}=0\Rightarrow24-12\text{x + 2}\sqrt{3}\text{ x}=0$
$\Rightarrow\text{x}=\frac{24}{12-2\sqrt{3}}\text{ OR }\frac{4(6+\sqrt{3})}{11}\text{m}$
$\therefore\text{y}=\frac{30-6\sqrt{3}}{11}\text{m}$
$\frac{\text{d}^{2}\text{A}}{\text{dx}^{2}}=(-12+3\sqrt{3})<0\therefore\text{Area is maximum for}$
$\text{x}=\frac{4(6+\sqrt{3})}{11}\text{m and y}=\frac{30-6\sqrt{3}}{11}\text{m}$.
 

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