The area (in sq. units) of the region bounded by the line 3x + y = 12, the lines y = 3 and y = 6, and the y-axis is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{15}{2}\)
From 3x + y = 12, \(x = \frac{12-y}{3}\), which is positive for y < 12. The region lies between y = 3 and y = 6 and to the right of the y-axis, so it is found by integrating with respect to y:
\( \displaystyle A = \int_{3}^{6}\frac{12-y}{3}\,dy \)
\( \displaystyle = \frac{1}{3}\left[12y-\frac{y^{2}}{2}\right]_{3}^{6} \)
\( \displaystyle = \frac{1}{3}\left[(72-18)-\left(36-\frac{9}{2}\right)\right] \)
\( \displaystyle = \frac{1}{3}\times\frac{45}{2} = \frac{15}{2} \)