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(D) The curves are $y=3x$,$y=\frac{27-3x}{2}$,and $y=3x-x\sqrt{x}$.First,find the intersection points:$1$. $y=3x$ and $y=3x-x\sqrt{x} \implies 3x=3x-x

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$162$

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(D) The curves are $y=3x$,$y=\frac{27-3x}{2}$,and $y=3x-x\sqrt{x}$.First,find the intersection points:$1$. $y=3x$ and $y=3x-x\sqrt{x} \implies 3x=3x-x\sqrt{x} \implies x\sqrt{x}=0 \implies x=0$. At $x=0$,$y=0$.$2$. $y=3x$ and $2y=27-3x \implies 6x=27-3x \implies 9x=27 \implies x=3$. At $x=3$,$y=9$.$3$. $y=3x-x\sqrt{x}$ and $2y=27-3x \implies 2(3x-x\sqrt{x})=27-3x \implies 6x-2x\sqrt{x}=27-3x \implies 9x-27=2x\sqrt{x}$. Squaring both sides: $(9(x-3))^2 = 4x^2(x) \implies 81(x-3)^2 = 4x^3$. Solving this gives $x=9$ as a root.The area $A$ is given by the integral:$A = \int_0^3 (3x - (3x-x\sqrt{x})) dx + \int_3^9 (\frac{27-3x}{2} - (3x-x\sqrt{x})) dx$$A = \int_0^3 x^{3/2} dx + \int_3^9 (\frac{27}{2} - \frac{9x}{2} + x^{3/2}) dx$$A = [\frac{2}{5}x^{5/2}]_0^3 + [\frac{27}{2}x - \frac{9}{4}x^2 + \frac{2}{5}x^{5/2}]_3^9$$A = (\frac{2}{5} \cdot 3^{5/2}) + ((\frac{27}{2} \cdot 9 - \frac{9}{4} \cdot 81 + \frac{2}{5} \cdot 9^{5/2}) - (\frac{27}{2} \cdot 3 - \frac{9}{4} \cdot 9 + \frac{2}{5} \cdot 3^{5/2}))$$A = \frac{2}{5} \cdot 3^{5/2} + (\frac{243}{2} - \frac{729}{4} + \frac{486}{5}) - (\frac{81}{2} - \frac{81}{4} + \frac{2}{5} \cdot 3^{5/2})$$A = \frac{243}{2} - \frac{729}{4} + \frac{486}{5} - \frac{81}{2} + \frac{81}{4} = \frac{162}{2} - \frac{648}{4} + \frac{486}{5} = 81 - 162 + 97.2 = 16.2$$10A = 162$.

Let the area of the region enclosed by the curves $y=3x$,$2y=27-3x$,and $y=3x-x\sqrt{x}$ be $A$. Then $10A$ is equal to

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