Question

If a rectangle is inscribed in an equilateral triangle of side length $2\sqrt{2}$ as shown in the figure, then the square of the largest area of such a rectangle is

Answer 1

Let the sides of rectangle be $x$ and $y.$
$\therefore x+\frac{2y}{\sqrt{3}}=2\sqrt{2}$
and area $\Delta =xy$
$\Rightarrow \Delta =y(2\sqrt{2}-\frac{2y}{\sqrt{3}})$
$\Rightarrow \Delta =(2\sqrt{2}y-\frac{2y^2}{\sqrt{3}})$
Differentiation w.r.t. $x,$ we get
$\frac{d\Delta }{dx}=(2\sqrt{2}-\frac{4y}{\sqrt{3}})$
For max/min $\frac{d\Delta }{dx}=0$
$\Rightarrow (2\sqrt{2}-\frac{4y}{\sqrt{3}})=0$
$\Rightarrow y=\frac{\sqrt{6}}{2}$
Now, $\frac{d^2\Delta }{d{x}^2}=(0-\frac{4}{\sqrt{3}})<0$
$\therefore$ area is maximum at $y=\frac{\sqrt{6}}{2}$
So, mximum area ${\Delta }_{max}=\sqrt{2}\times \frac{\sqrt{6}}{2}=\sqrt{3}$
Hence, ${\Delta }_{max}^2=3$