JEE MAIN - Mathematics (2025 - 24th January Morning Shift - No. 14)
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:
$\frac{821}{123}$
$\frac{567}{121}$
$\frac{730}{119}$
$\frac{625}{111}$
Explanation
$\mathrm{t} .\left(9-\frac{11 \mathrm{t}^2}{3}-\mathrm{t}\right)$
_24th_January_Morning_Shift_en_14_1.png)
$$\begin{aligned} & A=9 t-t^2-\frac{11}{3} t^3 \\ & \frac{d A}{d t}=9-2 t-11 t^2 \\ & \Rightarrow 11 t^2+2 t-9=0 \\ & 11 t^2+11 t-9 t-9=0 \\ & t=-1 \& t=\frac{9}{11} \end{aligned}$$
$\therefore \frac{\mathrm{dA}}{\mathrm{dt}}=$ _24th_January_Morning_Shift_en_14_2.png)
$$\begin{aligned} & \therefore \text { largest area }=\frac{9}{11}\left(9-\frac{11}{3}, \frac{81}{121}-\frac{9}{11}\right) \\ & =\frac{9}{11} \cdot \frac{63}{11}=\frac{567}{121} \end{aligned}$$
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