Let S = {(x, y) in R times R : x geq 0, y geq | vedclass

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(B) The region $S$ is bounded by $y^2 = 4x$,$y^2 = 12 - 2x$,and $3y + \sqrt{8}x = 5\sqrt{8}$.First,find the intersection points:For $y^2 = 4x$ and $y^

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$\frac{17}{3}$

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(B) The region $S$ is bounded by $y^2 = 4x$,$y^2 = 12 - 2x$,and $3y + \sqrt{8}x = 5\sqrt{8}$.First,find the intersection points:For $y^2 = 4x$ and $y^2 = 12 - 2x$,we have $4x = 12 - 2x \Rightarrow 6x = 12 \Rightarrow x = 2$. Then $y^2 = 8 \Rightarrow y = 2\sqrt{2}$.For $y^2 = 4x$ and $3y + 2\sqrt{2}x = 10\sqrt{2}$,substituting $x = y^2/4$ gives $3y + 2\sqrt{2}(y^2/4) = 10\sqrt{2} \Rightarrow 3y + \frac{y^2}{\sqrt{2}} = 10\sqrt{2} \Rightarrow y^2 + 3\sqrt{2}y - 20 = 0$.Solving this quadratic,$(y + 4\sqrt{2})(y - \sqrt{2}) = 0$. Since $y \geq 0$,$y = \sqrt{2}$. Then $x = 1/2$.However,looking at the region defined,the area is split at $x=2$. The area is $\int_0^2 \sqrt{4x} dx + \int_2^5 \sqrt{12-2x} dx$ is not correct based on the line constraint. The line $3y + 2\sqrt{2}x = 10\sqrt{2}$ passes through $(2, 2\sqrt{2})$ and $(5, 0)$.Area $= \int_0^2 2\sqrt{x} dx + 	ext{Area of triangle with vertices } (2,0), (5,0), (2, 2\sqrt{2})$.Area $= 2 \left[ \frac{x^{3/2}}{3/2} ight]_0^2 + \frac{1}{2} 	imes (5-2) 	imes 2\sqrt{2} = 2 	imes \frac{2}{3} 	imes 2\sqrt{2} + 3\sqrt{2} = \frac{8\sqrt{2}}{3} + 3\sqrt{2} = \frac{17\sqrt{2}}{3}$.Thus,$\alpha\sqrt{2} = \frac{17\sqrt{2}}{3} \Rightarrow \alpha = \frac{17}{3}$.

Let $S = \{(x, y) \in R \times R : x \geq 0, y \geq 0, y^2 \leq 4x, y^2 \leq 12 - 2x \text{ and } 3y + \sqrt{8}x \leq 5\sqrt{8}\}$. If the area of the region $S$ is $\alpha \sqrt{2}$,then $\alpha$ is equal to

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