Consider the region . The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is : [2025]

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Solution

Q.
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 : [2025]

1 $\frac{821}{123}$

2 $\frac{625}{111}$

3 $\frac{567}{121}$

4 $\frac{730}{119}$

Ans.
(3)

The given region R is shown below:

Here, x = t and $y = 9 - \frac{11 t^{3}}{3} - t$

Area of required rectangle,

$A = x y = t (9 - \frac{11 t^{2}}{3} - t) = 9 t - t^{2} - \frac{11}{3} t^{3}$

$\frac{d A}{d t} = 9 - 2 t - 11 t^{2}$

For critical points, $\frac{d A}{d t} = 0$

$\Rightarrow 11 t^{2} + 2 t - 9 = 0$

$\Rightarrow 11 t^{2} + 11 t - 9 t - 9 = 0$

$\Rightarrow t = - 1 or t = \frac{9}{11}$

$\frac{d^{2} A}{d t^{2}} = - 2 - 22 t$

$\frac{d^{2} A}{d t^{2}} > 0 at t = - 1$ i.e., minima and

$\frac{d^{2} A}{d t^{2}} < 0 at t = \frac{9}{11}$ i.e., maxima

$∴$ Maxima at $t = \frac{9}{11}$

$∴$ Largest area $= \frac{9}{11} (9 - \frac{11}{3} \times \frac{81}{121} - \frac{9}{11}) = \frac{9}{11} \times \frac{63}{11} = \frac{567}{121}$ .

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