Verified Solution • 2025–26 Session moderate QJEEM25Ma41c6: JEE Main 2025 Mathematics - Class 12 JEEM25Ma41c6 Solution: Consider the region . The area, of the largest rectangle of… Consider the region R={(x,y):x≤y≤9−113x2,x≥0}R=\left\{(x, y): x \leq y \leq 9-\frac{11}{3} x^2, x \geq 0\right\}R={(x,y):x≤y≤9−311x2,x≥0} . The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is: Multiple Choice Options A 821123\frac{821}{123}123821 B 567121\frac{567}{121}121567 C 730119\frac{730}{119}119730 D 625111\frac{625}{111}111625 Step-by-Step Verified Solution t.(9−11t23−t)\mathrm{t} .\left(9-\frac{11 \mathrm{t}^2}{3}-\mathrm{t}\right)t.(9−311t2−t) amp;A=9t−t2−113t3amp;dAdt=9−2t−11t2amp;⇒11t2+2t−9=0amp;11t2+11t−9t−9=0amp;t=−1&t=911\begin{aligned}& A=9 t-t^2-\frac{11}{3} t^3 \\& \frac{d A}{d t}=9-2 t-11 t^2 \\& \Rightarrow 11 t^2+2 t-9=0 \\& 11 t^2+11 t-9 t-9=0 \\& t=-1 \& t=\frac{9}{11}\end{aligned}amp;A=9t−t2−311t3amp;dtdA=9−2t−11t2amp;⇒11t2+2t−9=0amp;11t2+11t−9t−9=0amp;t=−1&t=119 ∴dAdt=\therefore \frac{\mathrm{dA}}{\mathrm{dt}}=∴dtdA= amp;∴ largest area =911(9−113,81121−911)amp;=911⋅6311=567121\begin{aligned}& \therefore \text { largest area }=\frac{9}{11}\left(9-\frac{11}{3}, \frac{81}{121}-\frac{9}{11}\right) \\& =\frac{9}{11} \cdot \frac{63}{11}=\frac{567}{121}\end{aligned}amp;∴ largest area =119(9−311,12181−119)amp;=119⋅1163=121567 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
