Question

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

• A. 28 – 30 loge 2
• B. 30 – 28 loge 2
• C. 30 – 32 loge 2
• D. 32 – 30 loge 2

Answer 1

Answer: (C)

30 – 32 loge 2

Answer 2

The given curves are:

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

From the second equation, solve for $y$:

$y = 6 - x$

Substitute $y = 6 - x$ into the first equation:

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

Simplifying the equation:

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

$2x - x^2 = -8$

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

Solving for $x$ by factoring:

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

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

The limits of integration are from $x = -2$ to $x = 4$. The area between the curves is given by the integral:

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

Solving the integral yields:

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