The area enclosed by the curves $$xy + 4y = 16$$ and $$x + y = 6$$ is equal to:

To find the area enclosed by the curves $$xy + 4y = 16$$ and $$x + y = 6$$, first rewrite the equations in terms of $$y$$.

The first curve: $$xy + 4y = 16$$ can be factored as $$y(x + 4) = 16$$, so $$y = \frac{16}{x+4}$$ for $$x \neq -4$$.

The second curve: $$x + y = 6$$ can be written as $$y = 6 - x$$.

Next, find the points of intersection by setting the $$y$$-values equal:

$$\frac{16}{x+4} = 6 - x$$

Multiply both sides by $$x+4$$ (assuming $$x \neq -4$$):

$$16 = (6 - x)(x + 4)$$

Expand the right side:

$$16 = 6x + 24 - x^2 - 4x$$

$$16 = -x^2 + 2x + 24$$

Bring all terms to one side:

$$x^2 - 2x - 8 = 0$$

Factor the quadratic:

$$(x - 4)(x + 2) = 0$$

So, $$x = 4$$ or $$x = -2$$.

Find the corresponding $$y$$-values using $$y = 6 - x$$:

When $$x = 4$$, $$y = 6 - 4 = 2$$, so point $$(4, 2)$$.

When $$x = -2$$, $$y = 6 - (-2) = 8$$, so point $$(-2, 8)$$.

The enclosed region is between $$x = -2$$ and $$x = 4$$. To determine which curve is above, evaluate at a test point, say $$x = 0$$:

For the hyperbola: $$y = \frac{16}{0+4} = 4$$.

For the line: $$y = 6 - 0 = 6$$.

Since $$6 > 4$$, the line $$y = 6 - x$$ is above the hyperbola $$y = \frac{16}{x+4}$$ in the interval $$[-2, 4]$$.

The area enclosed is given by the integral:

$$\text{Area} = \int_{-2}^{4} \left[ (6 - x) - \frac{16}{x+4} \right] dx$$

Split the integral into two parts:

$$\text{Area} = \int_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \frac{16}{x+4} dx$$

Compute the first integral:

Antiderivative of $$6 - x$$ is $$6x - \frac{x^2}{2}$$.

Evaluate from $$-2$$ to $$4$$:

At $$x = 4$$: $$6(4) - \frac{(4)^2}{2} = 24 - 8 = 16$$.

At $$x = -2$$: $$6(-2) - \frac{(-2)^2}{2} = -12 - 2 = -14$$.

So, $$\int_{-2}^{4} (6 - x) dx = 16 - (-14) = 30$$.

Now compute the second integral:

$$\int_{-2}^{4} \frac{16}{x+4} dx = 16 \int_{-2}^{4} \frac{1}{x+4} dx$$

Antiderivative of $$\frac{1}{x+4}$$ is $$\ln|x+4|$$.

Evaluate from $$-2$$ to $$4$$:

At $$x = 4$$: $$\ln|4+4| = \ln 8$$.

At $$x = -2$$: $$\ln|-2+4| = \ln 2$$.

So, $$\int_{-2}^{4} \frac{1}{x+4} dx = \ln 8 - \ln 2 = \ln \left( \frac{8}{2} \right) = \ln 4$$.

Thus, $$16 \int_{-2}^{4} \frac{1}{x+4} dx = 16 \ln 4$$.

Since $$\ln 4 = \ln (2^2) = 2 \ln 2$$, this becomes $$16 \times 2 \ln 2 = 32 \ln 2$$.

Therefore, the area is:

$$\text{Area} = 30 - 32 \ln 2$$

Since $$\log_e 2 = \ln 2$$, the area is $$30 - 32 \log_e 2$$.

Comparing with the options:

A. $$28 - 30\log_e 2$$
B. $$30 - 28\log_e 2$$
C. $$30 - 32\log_e 2$$
D. $$32 - 30\log_e 2$$

The expression matches option C.

The correct answer is C.