Question

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

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

Answer 1

The Correct Option is C

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:

\( \text{Area} = 30 - 32 \log 2 \)