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:

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

Answer 2

We need to find the area enclosed by the curves \( xy + 4y = 16 \) and \( x + y = 6 \).

Concept Used:

The area between two curves can be found by integrating the difference of the functions (top minus bottom) with respect to x, or by switching to y if more convenient. First, we find points of intersection, then set up the integral.

Step-by-Step Solution:

Step 1: Rewrite the equations in simpler forms.

Curve 1: \( xy + 4y = 16 \)
Factor y: \( y(x + 4) = 16 \) ⇒ \( y = \frac{16}{x+4} \)

Curve 2: \( x + y = 6 \) ⇒ \( y = 6 - x \)

Step 2: Find intersection points by solving \( \frac{16}{x+4} = 6 - x \).

\[ \frac{16}{x+4} = 6 - x \] \[ 16 = (6 - x)(x + 4) \] \[ 16 = 6x + 24 - x^2 - 4x \] \[ 16 = -x^2 + 2x + 24 \] \[ x^2 - 2x - 8 = 0 \] \[ (x - 4)(x + 2) = 0 \] \[ x = 4, \quad x = -2 \]

For \( x = 4 \), \( y = 6 - 4 = 2 \); for \( x = -2 \), \( y = 6 - (-2) = 8 \).
Intersection points: \( (4, 2) \) and \( (-2, 8) \).

Step 3: Determine which curve is above in the region.

Test point between x = -2 and x = 4, say x = 0:
Curve 1: \( y = \frac{16}{4} = 4 \)
Curve 2: \( y = 6 - 0 = 6 \)
So line \( y = 6 - x \) is above the curve \( y = \frac{16}{x+4} \) in the interval \([-2, 4]\).

Step 4: Set up the area integral.

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

Step 5: Compute the integral.

\[ \int_{-2}^{4} (6 - x) \, dx - \int_{-2}^{4} \frac{16}{x+4} \, dx \]

First part:

\[ \int_{-2}^{4} (6 - x) \, dx = \left[ 6x - \frac{x^2}{2} \right]_{-2}^{4} \] \[ = \left( 6(4) - \frac{16}{2} \right) - \left( 6(-2) - \frac{4}{2} \right) \] \[ = (24 - 8) - (-12 - 2) = 16 - (-14) = 30 \]

Second part:

\[ \int_{-2}^{4} \frac{16}{x+4} \, dx = 16 \left[ \ln|x+4| \right]_{-2}^{4} \] \[ = 16 \left( \ln 8 - \ln 2 \right) = 16 \ln \left( \frac{8}{2} \right) = 16 \ln 4 = 16 \cdot 2 \ln 2 = 32 \ln 2 \]

Step 6: Combine results.

\[ \text{Area} = 30 - 32 \ln 2 \]

Therefore, the area enclosed by the curves is \( \mathbf{30 - 32 \ln 2} \).