Consider the region R = {(x, y) : x leq y leq | vedclass

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(C) Let the rectangle have vertices at $(t, t)$,$(t, 9 - \frac{11}{3}t^2)$,$(0, 9 - \frac{11}{3}t^2)$,and $(0, t)$.The width of the rectangle is $t$ a

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$\frac{567}{121}$

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(C) Let the rectangle have vertices at $(t, t)$,$(t, 9 - \frac{11}{3}t^2)$,$(0, 9 - \frac{11}{3}t^2)$,and $(0, t)$.The width of the rectangle is $t$ and the height is $(9 - \frac{11}{3}t^2 - t)$.The area $A$ of the rectangle is given by $A(t) = t \cdot (9 - \frac{11}{3}t^2 - t) = 9t - t^2 - \frac{11}{3}t^3$.To find the maximum area,we differentiate $A(t)$ with respect to $t$:$\frac{dA}{dt} = 9 - 2t - 11t^2$.Setting $\frac{dA}{dt} = 0$,we get $11t^2 + 2t - 9 = 0$.Factoring the quadratic equation: $11t^2 + 11t - 9t - 9 = 0 \Rightarrow 11t(t + 1) - 9(t + 1) = 0 \Rightarrow (11t - 9)(t + 1) = 0$.Since $x \geq 0$,we have $t = \frac{9}{11}$.The maximum area is $A(\frac{9}{11}) = \frac{9}{11} \cdot (9 - \frac{11}{3} \cdot (\frac{9}{11})^2 - \frac{9}{11}) = \frac{9}{11} \cdot (9 - \frac{27}{11} - \frac{9}{11}) = \frac{9}{11} \cdot (9 - \frac{36}{11}) = \frac{9}{11} \cdot (\frac{99 - 36}{11}) = \frac{9}{11} \cdot \frac{63}{11} = \frac{567}{121}$.

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 with sides parallel to the coordinate axes and inscribed in $R$ is:

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