JEE Main 2025 Jan 24 Shift 1, Mathematics Q12: Maxima & Minima

The region sits between the line $y=x$ (below) and the parabola $y=9-\frac{11}{3}{x}^2$ (above), with $x\ge 0$.

An axis-aligned rectangle inscribed in $R$ with one corner on $y=x$ at $(t,t)$ and the opposite corner on the parabola has dimensions:

width $=t$ (from $x=0$ to $x=t$), height $=\left(9-\frac{11}{3}{t}^2\right)-t$.

Area $A(t)=t\left(9-\frac{11}{3}{t}^2-t\right)=9t-\frac{11}{3}{t}^3-t^2$.

$A^{\prime }(t)=9-11t^2-2t=0\Rightarrow 11t^2+2t-9=0$

Factor: $(11t-9)(t+1)=0\Rightarrow t=\frac{9}{11}$ (positive root).

At $t=\frac{9}{11}$:

Height $=9-\frac{11}{3}\cdot \frac{81}{121}-\frac{9}{11}=9-\frac{27}{11}-\frac{9}{11}=\frac{99-36}{11}=\frac{63}{11}$

Area $=\frac{9}{11}\cdot \frac{63}{11}=\frac{567}{121}$.

$\therefore$ Option (3).