JEE Main 2024 — 8 April, Shift 2 — Question 19

The parabola is $y^2 = 2x$ so $x = \frac{y^2}{2}$.

The region is bounded on the left by the parabola and on the right by $x = 24$, symmetric about the x-axis. Consider a rectangle inscribed with sides parallel to axes, right side on $x = 24$ and left side on parabola at $y$ (with $y > 0$). Width = $24 - \frac{y^2}{2}$, height = $2y$. Area $A(y) = 2y\left(24 - \frac{y^2}{2}\right) = 48y - y^3$. Differentiate: $\frac{dA}{dy} = 48 - 3y^2$. Set $\frac{dA}{dy} = 0 \Rightarrow 48 - 3y^2 = 0 \Rightarrow y^2 = 16 \Rightarrow y = 4$ (positive). Second derivative $\frac{d^2A}{dy^2} = -6y < 0$ at $y = 4$, confirming maximum. Maximum area $A_{\text{max}} = 48(4) - (4)^3 = 192 - 64 = 128$.