The area (in sq. units) of the region \(\{(x, y): x \ge 0,\ y \ge 0,\ 2x+y \le 8,\ x+2y \le 8\}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{32}{3}\)
The lines 2x + y = 8 and x + 2y = 8 meet where
\( \displaystyle x = y = \frac{8}{3} \)
The region lies under both lines, so its top is the lower of the two. For \(0 \le x \le \frac{8}{3}\) the lower line is \(y = \displaystyle\frac{8-x}{2}\); for \(\frac{8}{3} \le x \le 4\) it is y = 8 − 2x, which meets the x-axis at x = 4.
\( \displaystyle \int_{0}^{8/3}\frac{8-x}{2}\,dx = \left[4x-\frac{x^{2}}{4}\right]_{0}^{8/3} \)
\( \displaystyle = \frac{32}{3}-\frac{16}{9} = \frac{80}{9} \)
\( \displaystyle \int_{8/3}^{4}(8-2x)\,dx = \left[8x-x^{2}\right]_{8/3}^{4} \)
\( \displaystyle = 16-\frac{128}{9} = \frac{16}{9} \)
\( \displaystyle A = \frac{80}{9}+\frac{16}{9} = \frac{96}{9} = \frac{32}{3} \)
Why the other options are incorrect
Option (1) 16 — Incorrect. This is the triangle cut off by one line alone, for example \(\frac{1}{2}\times4\times8\) for 2x + y ≤ 8; the other condition removes part of it.
Option (2) 8 — Incorrect. The two parts give \(\frac{80}{9}+\frac{16}{9} = \frac{32}{3}\), which is more than 8.
Option (4) \(\displaystyle\frac{40}{3}\) — Incorrect. The two parts add up to \(\frac{80}{9}+\frac{16}{9} = \frac{32}{3}\).