The area bounded by `xy+4y=16` and `x+y=6` is

Options

A. `32-42ln3`
B. `42-32ln2`
C. `30-32ln2`
D. `33+16ln3`

Answer

To find the area bounded by the curves defined by the equations \( xy + 4y = 16 \) and \( x + y = 6 \), we will follow these steps:

### Step 1: Rewrite the equations
First, let's rewrite the equations in a more manageable form.

1. The first equation \( xy + 4y = 16 \) can be rearranged to express \( y \):
\[ y(x + 4) = 16 \implies y = \frac{16}{x + 4} \]

2. The second equation \( x + y = 6 \) can also be rearranged to express \( y \):
\[ y = 6 - x \]

### Step 2: Find the points of intersection
Next, we need to find the points where these two curves intersect. We set the two expressions for \( y \) equal to each other:
\[ \frac{16}{x + 4} = 6 - x \]

To eliminate the fraction, we can multiply both sides by \( x + 4 \):
\[ 16 = (6 - x)(x + 4) \]

Expanding the right side:
\[ 16 = 6x + 24 - x^2 - 4x \]
\[ 16 = -x^2 + 2x + 24 \]

Rearranging gives us:
\[ x^2 - 2x + 8 = 0 \]

### Step 3: Solve the quadratic equation
Now we can solve the quadratic equation \( x^2 - 2x + 8 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 - 32}}{2} = \frac{2 \pm \sqrt{-28}}{2} \]
Since the discriminant is negative, there are no real solutions, meaning the curves do not intersect.

### Step 4: Determine the area between the curves
Since the curves do not intersect, we can find the area between them by integrating from the leftmost point of the line to the rightmost point of the line.

1. The line \( x + y = 6 \) intersects the axes at \( (6, 0) \) and \( (0, 6) \).
2. The curve \( y = \frac{16}{x + 4} \) approaches the x-axis as \( x \) increases.

### Step 5: Set up the integral
To find the area between the two curves from \( x = 0 \) to \( x = 6 \):
\[ \text{Area} = \int_0^6 \left( (6 - x) - \frac{16}{x + 4} \right) dx \]

### Step 6: Calculate the integral
Now we can calculate the integral:
\[ \text{Area} = \int_0^6 (6 - x) \, dx - \int_0^6 \frac{16}{x + 4} \, dx \]

Calculating the first integral:
\[ \int_0^6 (6 - x) \, dx = \left[ 6x - \frac{x^2}{2} \right]_0^6 = \left[ 36 - 18 \right] - 0 = 18 \]

Calculating the second integral:
\[ \int_0^6 \frac{16}{x + 4} \, dx = 16 \left[ \ln|x + 4| \right]_0^6 = 16 \left[ \ln(10) - \ln(4) \right] = 16 \ln\left(\frac{10}{4}\right) = 16 \ln(2.5) \]

### Step 7: Final area calculation
Thus, the total area is:
\[ \text{Area} = 18 - 16 \ln(2.5) \]

### Conclusion
The area bounded by the curves \( xy + 4y = 16 \) and \( x + y = 6 \) is:
\[ \text{Area} = 18 - 16 \ln(2.5) \]