JEE Main 2025 — 24 January, Morning Shift — Question 12

$\text{Region\; }R=\{(x,y):\;x\ge0,\;x\le y\le 9-\tfrac{11}{3}x^2\}.$

$\text{Any\; axis–parallel \; rectangle\; inscribed\; in\; }R\; \text{ has\; for\; }x\; \in[u,v]\;(0\le u < v)\text{ constant\; }y\;\text{-interval\; }[y_1,y_2]$

$\text{For\; the\; rectangle\; to\; lie\; in\; }R,\text{ we\; need\; }y_1\ge\max_{x\in[u,v]}x=v,\quad y_2\le\min_{x\in[u,v]}\Big(9-\tfrac{11}{3}x^2\Big)=9-\tfrac{11}{3}v^2.$

$\text{Thus\; the\; maximal\; height\; for\; a\; given\; }[u,v]\text{ is }H(v)=9-\tfrac{11}{3}v^2-v,$

$\text{and\; the\; width\; is\; }W=v-u\le v\ (\text{maximized\; at\; }u=0).$

$\text{Hence\; it\; suffices\; to\; take\; }u=0\text{ and\; maximize\; }A(v)=v\Big(9-\;v-\tfrac{11}{3}v^2\Big),\;0\le v\le v_{\max}.$

$A(v)=9v-v^2-\tfrac{11}{3}v^3\quad\Rightarrow\quad A'(v)=9-2v-11v^2.$

$A'(v)=0\iff 11v^2+2v-9=0\iff v=\frac{-2\pm\sqrt{4+396}}{22}=\frac{-2\pm20}{22}.$

$\text{Positive\; root:\; }v=\frac{18}{22}=\frac{9}{11}.$

$\text{Maximum\; area:\; }A\!\left(\tfrac{9}{11}\right)=9\cdot\frac{9}{11}-\left(\frac{9}{11}\right)^2-\frac{11}{3}\left(\frac{9}{11}\right)^3.$

$9\cdot\frac{9}{11}=\frac{81}{11}=\frac{891}{121},\quad \left(\frac{9}{11}\right)^2=\frac{81}{121},\quad \frac{11}{3}\left(\frac{9}{11}\right)^3=\frac{243}{121}.$

$A\!\left(\tfrac{9}{11}\right)=\frac{891}{121}-\frac{81}{121}-\frac{243}{121}=\frac{567}{121}.$

$\boxed{\dfrac{567}{121}}$