1. Write the curves in convenient form

The given curves are:

$xy+4y=16$ $x+y=6$

From the first equation, $y(x+4)=16 \implies y=\frac{16}{x+4}$

From the second equation, $y=6-x$

So we need the area enclosed between $y=\frac{16}{x+4} \quad \text{and} \quad y=6-x.$

2. Find the points of intersection

Set the two expressions for $y$ equal:

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

Multiply by $(x+4)$:

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

Expand:

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

Rearrange:

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

Factor:

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

Hence, $x=4,\,-2$

Now find corresponding $y$ values using $y=6-x$:

• For $x=4$, $y=2$
• For $x=-2$, $y=8$

So the curves intersect at $(4,2)$ and $(-2,8)$.

3. Determine which curve is above the other

Take a test value between $x=-2$ and $x=4$, say $x=0$.

Then $y_{\text{line}}=6-0=6$ $y_{\text{curve}}=\frac{16}{0+4}=4$

So on $[-2,4]$, the line $y=6-x$ lies above the curve $y=\dfrac{16}{x+4}$.

Therefore area is

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

4. Evaluate the integral

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

First integral:

$\int (6-x)dx=6x-\frac{x^2}{2}$

So, $\left[6x-\frac{x^2}{2}\right]_{-2}^{4}$

At $x=4$: $24-8=16$

At $x=-2$: $-12-2=-14$

Thus, $16-(-14)=30$

Now second integral:

$\int \frac{16}{x+4}dx=16\ln|x+4|$

So, $\left[16\ln|x+4|\right]_{-2}^{4}=16(\ln 8-\ln 2)=16\ln 4=32\ln 2$

Hence, $A=30-32\ln 2$

5. Match with the options

$A=30-32\log_e 2$

This corresponds to Option C.

6. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored correct answer.