If area of region enclosed by `y = 3x, 2y = 27 – 3x` and `y = 3x – x^(3/2)` is A, then 10 A is equal to

Options

A. `(80)/7`
B. `(81)/5`
C. `(101)/5`
D. `(91)/5`

Answer

To find the area \( A \) enclosed by the curves \( y = 3x \), \( 2y = 27 - 3x \), and \( y = 3x - x^{3/2} \), we will follow these steps:

### Step 1: Find the intersection points of the curves

1. **Intersection of \( y = 3x \) and \( 2y = 27 - 3x \)**:
- Substitute \( y = 3x \) into \( 2y = 27 - 3x \):
\[ 2(3x) = 27 - 3x \implies 6x = 27 - 3x \implies 9x = 27 \implies x = 3 \]
- Substitute \( x = 3 \) back into \( y = 3x \):
\[ y = 3(3) = 9 \]
- Intersection point: \( (3, 9) \).

2. **Intersection of \( y = 3x \) and \( y = 3x - x^{3/2} \)**:
- Set \( 3x = 3x - x^{3/2} \):
\[ 0 = -x^{3/2} \implies x = 0 \]
- Intersection point: \( (0, 0) \).

3. **Intersection of \( 2y = 27 - 3x \) and \( y = 3x - x^{3/2} \)**:
- First, express \( y \) from \( 2y = 27 - 3x \):
\[ y = \frac{27 - 3x}{2} \]
- Set this equal to \( 3x - x^{3/2} \):
\[ \frac{27 - 3x}{2} = 3x - x^{3/2} \]
- Multiply through by 2 to eliminate the fraction:
\[ 27 - 3x = 6x - 2x^{3/2} \implies 2x^{3/2} - 9x + 27 = 0 \]
- This is a cubic equation in \( x \). We can find the roots using numerical or graphical methods, but we know from the previous intersections that \( x = 9 \) is a solution (as we will see in the next steps).

### Step 2: Set up the area integrals

The area \( A \) can be split into two parts:
1. From \( x = 0 \) to \( x = 3 \): Area between \( y = 3x \) and \( y = 3x - x^{3/2} \).
2. From \( x = 3 \) to \( x = 9 \): Area between \( y = \frac{27 - 3x}{2} \) and \( y = 3x - x^{3/2} \).

#### Area from \( x = 0 \) to \( x = 3 \):
\[ A_1 = \int_0^3 \left( (3x) - (3x - x^{3/2}) \right) \, dx = \int_0^3 x^{3/2} \, dx \]

#### Area from \( x = 3 \) to \( x = 9 \):
\[ A_2 = \int_3^9 \left( \frac{27 - 3x}{2} - (3x - x^{3/2}) \right) \, dx \]

### Step 3: Calculate the integrals

1. **Calculate \( A_1 \)**:
\[ A_1 = \int_0^3 x^{3/2} \, dx = \left[ \frac{2}{5} x^{5/2} \right]_0^3 = \frac{2}{5} (3^{5/2}) = \frac{2}{5} \cdot 27 = \frac{54}{5} \]

2. **Calculate \( A_2 \)**:
\[ A_2 = \int_3^9 \left( \frac{27 - 3x}{2} - (3x - x^{3/2}) \right) \, dx \]
Simplifying the integrand:
\[ = \int_3^9 \left( \frac{27 - 3x - 6x + 2x^{3/2}}{2} \right) \, dx = \int_3^9 \left( \frac{27 - 9x + 2x^{3/2}}{2} \right) \, dx \]
Now integrate:
\[ A_2 = \frac{1}{2} \left[ 27x - \frac{9}{2} x^2 + \frac{4}{5} x^{5/2} \right]_3^9 \]
Evaluate the limits to find \( A_2 \).

### Step 4: Combine areas and calculate \( 10A \)

Finally, combine \( A_1 \) and \( A_2 \) to find \( A \):
\[ A = A_1 + A_2 \]
Then calculate \( 10A \).