Question:

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

Updated On: Jan 1, 2025
  • 28 – 30 loge 2
  • 30 – 28 loge 2
  • 30 – 32 loge 2
  • 32 – 30 loge 2
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The Correct Option is C

Solution and Explanation

The given curves are:

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

From the second equation, solve for y y :

y=6x y = 6 - x

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

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

Simplifying the equation:

6xx2+244x=16 6x - x^2 + 24 - 4x = 16

2xx2=8 2x - x^2 = -8

x22x8=0 x^2 - 2x - 8 = 0

Solving for x x by factoring:

(x4)(x+2)=0 (x - 4)(x + 2) = 0

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

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

Area=24(16x+4(6x))dx \text{Area} = \int_{-2}^{4} \left( \frac{16}{x + 4} - (6 - x) \right) \, dx

Solving the integral yields:

Area=3032log2 \text{Area} = 30 - 32 \log 2

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