JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Application of Derivatives

Let the equilateral triangle have side length $a=2\sqrt{2}$. Place the base along the x-axis with one vertex at the origin. The height of the equilateral triangle is $h=\frac{\sqrt{3}}{2}a=\sqrt{6}$.

Let the rectangle have width $w$ along the base and height $y$. By similar triangles, the relationship between $y$ and $w$ is: $y=h\left(1-\frac{w}{a}\right)=\sqrt{6}\left(1-\frac{w}{2\sqrt{2}}\right)$.

The area of the rectangle is $A=w\cdot y=w\sqrt{6}\left(1-\frac{w}{2\sqrt{2}}\right)$.

To maximize, differentiate with respect to $w$ and set to zero:
$\frac{dA}{dw}=\sqrt{6}\left(1-\frac{w}{\sqrt{2}}\right)=0$

This gives $w=\sqrt{2}$, and $y=\sqrt{6}\left(1-\frac{1}{2}\right)=\frac{\sqrt{6}}{2}$.

Maximum area: $A_{\text{max}}=\sqrt{2}\cdot \frac{\sqrt{6}}{2}=\frac{\sqrt{12}}{2}=\sqrt{3}$.

The square of the largest area is $(\sqrt{3}{)}^2=3$.