The maximum value of a LPP $z = 3x + 4y$ subject to the constraints: $x + y ≤6, x≥0,y≥0$ is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 24
$\text{Maximize } z=3x+4y \text{ subject to } x+y\le 6,\; x\ge 0,\; y\ge 0.$
Corner points: $(0,0),\ (6,0),\ (0,6)$.
$z(0,0)=0,\quad z(6,0)=18,\quad z(0,6)=24.$
Maximum value = 24 at $(0,6)$.
