AOD – Maxima – Minima – JEE Main 24 January 2025 Shift 1

Solution:

The region is bounded by the y-axis ($x \ge 0$), the line $y=x$, and the parabola $y = 9 – \frac{11}{3}x^2$.
Let the rectangle have width $t$ along the x-axis. Since it is inscribed with sides parallel to axes and $x \ge 0$, the rectangle spans from $x=0$ to $x=t$.
The vertices of the right side of the rectangle lie on the boundaries.
The bottom-right vertex lies on the line $y=x$, so its coordinates are $(t, t)$.
The top-right vertex lies on the parabola, so its coordinates are $(t, 9 – \frac{11}{3}t^2)$.


The height of the rectangle is the difference in y-coordinates:
$$ h(t) = \left(9 – \frac{11}{3}t^2\right) – t $$
The width of the rectangle is $t$.
$$ \text{Area } A(t) = t \times h(t) = t\left(9 – \frac{11}{3}t^2 – t\right) $$
$$ A(t) = 9t – \frac{11}{3}t^3 – t^2 $$

To maximize area, find $\frac{dA}{dt}$:
$$ \frac{dA}{dt} = 9 – 11t^2 – 2t $$
Set derivative to zero:
$$ 11t^2 + 2t – 9 = 0 $$
$$ 11t^2 + 11t – 9t – 9 = 0 $$
$$ 11t(t+1) – 9(t+1) = 0 $$
$$ (11t – 9)(t + 1) = 0 $$
Since $t > 0$, we have $t = \frac{9}{11}$.

Calculate the maximum area at $t = \frac{9}{11}$:
$$ A\left(\frac{9}{11}\right) = \frac{9}{11} \left( 9 – \frac{11}{3}\left(\frac{9}{11}\right)^2 – \frac{9}{11} \right) $$
$$ = \frac{9}{11} \left( 9 – \frac{11}{3} \cdot \frac{81}{121} – \frac{9}{11} \right) $$
$$ = \frac{9}{11} \left( 9 – \frac{27}{11} – \frac{9}{11} \right) $$
$$ = \frac{9}{11} \left( 9 – \frac{36}{11} \right) = \frac{9}{11} \left( \frac{99 – 36}{11} \right) $$
$$ = \frac{9}{11} \left( \frac{63}{11} \right) = \frac{567}{121} $$

Ans. (3)