The minimum value of $z=3 x+6 y$ subject to the constraints $2 x+3 y ≤ 180, x+y ≥ 60, x ≥ 3 y, x ≥ 0, y ≥ 0$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - 180
$\text{Minimise } Z = 3x + 6y$
$2x + 3y \le 180,\;\; x + y \ge 60,\;\; x \ge 3y,\;\; x,y \ge 0$
$\text{Corner points from intersections:}$
$2x+3y=180 \text{ and } x+y=60 \Rightarrow x=0,\; y=60$
$x+y=60 \text{ and } x=3y \Rightarrow 3y+y=60 \Rightarrow y=15,\; x=45$
$2x+3y=180 \text{ and } x=3y \Rightarrow 6y+3y=180 \Rightarrow y=20,\; x=60$
$y=0,\; x+y=60 \Rightarrow (60,0)$
$\text{Evaluate } Z:$
$Z(0,60)=360,\;\; Z(45,15)=225,\;\; Z(60,20)=300,\;\; Z(60,0)=180$
$\text{Minimum } Z = 180$
The minimum value is $180$.