Write the curves in simpler form

The given curves are: $y=3x$ $2y=27-3x \Rightarrow y=\frac{27-3x}{2}$ $y=3x-x\sqrt{x}$

Find the points of intersection

We identify which curves bound the enclosed region.

(i) Intersection of $y=3x$ and $y=\dfrac{27-3x}{2}$

$3x=\frac{27-3x}{2}$ $6x=27-3x$ $9x=27$ $x=3$ Then $y=3x=9$ So one point is $(3,9)$.

(ii) Intersection of $y=3x$ and $y=3x-x\sqrt{x}$

$3x=3x-x\sqrt{x}$ $x\sqrt{x}=0$ $x=0$ So the point is $(0,0)$.

(iii) Intersection of $y=\dfrac{27-3x}{2}$ and $y=3x-x\sqrt{x}$

$3x-x\sqrt{x}=\frac{27-3x}{2}$ $6x-2x\sqrt{x}=27-3x$ $9x-2x\sqrt{x}=27$

Check $x=9$: $9(9)-2(9)(3)=81-54=27$ so $x=9$ works.

Then $y=3(9)-9\cdot 3=27-27=0$ So the point is $(9,0)$.

Thus the enclosed region has vertices: $(0,0),\ (3,9),\ (9,0).$

Determine the upper and lower curves in each interval

The curve $y=3x-x\sqrt{x}$ lies below $y=3x$ for $x>0$.

Also, the line $y=\dfrac{27-3x}{2}$ intersects $y=3x$ at $x=3$.

So the enclosed area is split as:

• For $0\le x\le 3$: upper curve is $y=3x$, lower curve is $y=3x-x\sqrt{x}$.
• For $3\le x\le 9$: upper curve is $y=\dfrac{27-3x}{2}$, lower curve is $y=3x-x\sqrt{x}$.

Set up the area integral

Therefore, $A=\int_0^3 \left[3x-(3x-x\sqrt{x})\right]dx + \int_3^9 \left[\frac{27-3x}{2}-(3x-x\sqrt{x})\right]dx$

Simplify: $A=\int_0^3 x\sqrt{x}\,dx + \int_3^9 \left(\frac{27-9x}{2}+x\sqrt{x}\right)dx$

Evaluate the first integral

$\int_0^3 x\sqrt{x}\,dx=\int_0^3 x^{3/2}\,dx$ $=\left[\frac{2}{5}x^{5/2}\right]_0^3$ $=\frac{2}{5}\cdot 3^{5/2}$ $=\frac{2}{5}\cdot 9\sqrt{3}$ $=\frac{18\sqrt{3}}{5}$

Evaluate the second integral

Add both parts

$A=\frac{18\sqrt3}{5}+\left(\frac{81}{5}-\frac{18\sqrt3}{5}\right)$ $A=\frac{81}{5}$

Compute $10A$

$10A=10\cdot \frac{81}{5}=162$

Match with options

$10A=162$ So the correct option is C.