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:

Question

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:

Aicharya · Accepted Answer

$\mathrm{t} .\left(9-\frac{11 \mathrm{t}^2}{3}-\mathrm{t}\right)$

$\begin{aligned} & \mathrm{A}=9 \mathrm{t}-\mathrm{t}^2-\frac{11}{3} \mathrm{t}^3 \\ & \frac{\mathrm{dA}}{\mathrm{dt}}=9-2 \mathrm{t}-11 \mathrm{t}^2 \\ & \Rightarrow 11 \mathrm{t}^2+2 \mathrm{t}-9=0 \\ & 11 \mathrm{t}^2+11 \mathrm{t}-9 \mathrm{t}-9=0 \\ & \mathrm{t}=-1 \& \mathrm{t}=\frac{9}{11}\end{aligned}$

$\begin{aligned} & \therefore \text { largest area }=\frac{9}{11}\left(9-\frac{11}{3}, \frac{81}{121}-\frac{9}{11}\right) \\ & =\frac{9}{11} \cdot \frac{63}{11}=\frac{567}{121}\end{aligned}$

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…

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