Consider the region $R=\left\{(x, y): x \leq y \leq 9-\frac{11}{3} x^{2}, x \geq 0\right\}$. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is :

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Application of Derivatives

Consider the region

R = {(x, y) : x \leq y \leq 9 - \frac{11}{3} x^{2}, x \geq 0}

. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is :

Answer: (C)

Take a point on the line

y = x

as one corner of the rectangle, say

(t, t)

, and the point on the curve

y = 9 - \frac{11}{3} x^{2}

directly above it as the opposite corner. The rectangle then runs from

x = 0

x = t

with height

(9 - \frac{11}{3} t^{2}) - t

, so its area is

A (t) = t [(9 - \frac{11}{3} t^{2}) - t] = 9 t - t^{2} - \frac{11}{3} t^{3} .

A^{'} (t) = 9 - 2 t - 11 t^{2} = 0 \Rightarrow 11 t^{2} + 2 t - 9 = 0 \Rightarrow (11 t - 9) (t + 1) = 0.

Since

t \geq 0

t = \frac{9}{11}

, which gives a maximum (

A^{'} (t)

is positive before this root and negative after it).

A (\frac{9}{11}) = \frac{9}{11} (9 - \frac{11}{3} \cdot \frac{81}{121} - \frac{9}{11}) = \frac{9}{11} (\frac{99 - 27 - 9}{11}) = \frac{9}{11} \cdot \frac{63}{11} = \frac{567}{121} .

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option