A is the area of region `0leylemin(2x,6x-x^2)`, then find 12A

Options

A. 304
B. 302
C. 288
D. 312

Answer

To solve the problem of finding the area \( A \) of the region defined by \( 0 \leq y \leq \min(2x, 6x - x^2) \), and then calculating \( 12A \), we will follow these steps:

### Step 1: Identify the curves
We have two functions:
1. \( y = 2x \) (a straight line)
2. \( y = 6x - x^2 \) (a downward-opening parabola)

### Step 2: Find the points of intersection
To find the area between these curves, we first need to determine where they intersect. We set the equations equal to each other:

\[ 2x = 6x - x^2 \]

Rearranging gives:

\[ x^2 - 4x = 0 \]

Factoring out \( x \):

\[ x(x - 4) = 0 \]

This gives us the solutions:

\[ x = 0 \quad \text{and} \quad x = 4 \]

### Step 3: Find the corresponding y-values
Now, we can find the y-values at these x-values:
- At \( x = 0 \):
\[ y = 2(0) = 0 \]
- At \( x = 4 \):
\[ y = 2(4) = 8 \]

So, the points of intersection are \( (0, 0) \) and \( (4, 8) \).

### Step 4: Set up the area integral
The area \( A \) between the curves from \( x = 0 \) to \( x = 4 \) can be calculated using the integral:

\[ A = \int_0^4 (6x - x^2 - 2x) \, dx \]

This simplifies to:

\[ A = \int_0^4 (4x - x^2) \, dx \]

### Step 5: Calculate the integral
Now we compute the integral:

\[ A = \int_0^4 (4x - x^2) \, dx = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 \]

Calculating the upper limit:

\[ = 2(4^2) - \frac{(4^3)}{3} = 2(16) - \frac{64}{3} = 32 - \frac{64}{3} \]

To combine these, convert \( 32 \) to a fraction:

\[ 32 = \frac{96}{3} \]

Thus,

\[ A = \frac{96}{3} - \frac{64}{3} = \frac{32}{3} \]

### Step 6: Calculate \( 12A \)
Now we find \( 12A \):

\[ 12A = 12 \times \frac{32}{3} = \frac{384}{3} = 128 \]

### Final Answer
Thus, the final answer is:

\[ \boxed{128} \]