Consider the region $$R=\{(x,y): x \le y \le 9-\tfrac{11}{3}x^2,\; x\ge 0\}.$$  The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in  $$R$$ is:

Find the area of the largest rectangle with sides parallel to axes inscribed in region $$R = \{(x,y): x \leq y \leq 9 - \frac{11}{3}x^2, x \geq 0\}$$.

We begin by considering a rectangle whose vertices are $$(0,y_1),\,(x,y_1),\,(x,y_2),\,(0,y_2)$$ subject to $$x\le y_1$$ and $$y_2\le 9-\frac{11}{3}x^2\,. $$

To maximize the area, it is natural to take the lower bound at $$y_1=x$$ and the upper bound at $$y_2=9-\frac{11}{3}x^2\,. $$

Area = $$x\bigl(9 - \frac{11}{3}x^2 - x\bigr)$$

$$= 9x - \frac{11}{3}x^3 - x^2$$

Next, we set the derivative of the area with respect to $$x$$ equal to zero:

$$\frac{dA}{dx} = 9 - 11x^2 - 2x = 0$$

$$11x^2 + 2x - 9 = 0$$

$$x = \frac{-2 \pm \sqrt{4+396}}{22} = \frac{-2 \pm 20}{22}$$

$$x = \frac{18}{22} = \frac{9}{11}$$ (taking the positive root)

Substituting $$x=\frac{9}{11}$$ back into the area formula gives

$$A = 9 \cdot \frac{9}{11} - \frac{11}{3}\left(\frac{9}{11}\right)^3 - \left(\frac{9}{11}\right)^2$$

$$= \frac{81}{11} - \frac{11}{3} \cdot \frac{729}{1331} - \frac{81}{121}$$

$$= \frac{81}{11} - \frac{729}{363} - \frac{81}{121}$$

$$= \frac{81}{11} - \frac{243}{121} - \frac{81}{121}$$

$$= \frac{81 \times 11}{121} - \frac{324}{121}$$

$$= \frac{891 - 324}{121} = \frac{567}{121}$$

The correct answer is Option 4: $$\frac{567}{121}$$.