The corner points of the feasible region determined by the following system of linear inequalities
\( x+2y\le 120,\ x+y\ge 60,\ x-2y\ge 0,\ x,y\ge 0 \) are (60,0) , (120, 0), (60, 30) and (40, 20). Let
\( Z=ax+by,\ a,b\gt 0 \). If the maximum Z at (120,0) and (60, 30) and minimum Z is 300 at (60, 0), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3a+2b=35 \)
The maximum occurs at both (120, 0) and (60, 30), so \( Z \) has the same value at these two points:
\( \displaystyle 120a=60a+30b \)
\( \displaystyle b=2a \)
The minimum value 300 occurs at (60, 0):
\( \displaystyle 60a=300 \)
\( \displaystyle a=5,\qquad b=10 \)
Now \( 3a+2b=15+20=35 \), which is Option (2).