The feasible region of a maximisation problem whose objective function is \( z=3x+5y \) is shown as shaded region below. Then the constraints are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( x\ge 0,\ y\ge 0,\ x+2y\le 8,\ 3x+2y\le 12 \)
The line through the intercepts (8, 0) and (0, 4) is
\( \displaystyle x+2y=8 \)
The line through the intercepts (4, 0) and (0, 6) is
\( \displaystyle 3x+2y=12 \)
The shaded region is in the first quadrant, so \( x\ge 0 \) and \( y\ge 0 \).
The origin lies in the shaded region, and at (0, 0) we get \( 0\le 8 \) and \( 0\le 12 \). So both lines take “≤”.