The maximum value of \( z=4x+2y \), subjected to the constraints:
\( 2x+3y\le 18,\ x+y\ge 5 \) ;
\( x,y\ge 0 \), is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 36
The two boundary lines meet the axes at:
\( \displaystyle 2x+3y=18:\ (9,0)\ \text{and}\ (0,6) \)
\( \displaystyle x+y=5:\ (5,0)\ \text{and}\ (0,5) \)
The feasible region lies below \( 2x+3y=18 \), above \( x+y=5 \) and in the first quadrant. The two lines do not meet in the first quadrant (solving them gives x = −3), so the region is the quadrilateral with corners (5, 0), (9, 0), (0, 6) and (0, 5).
Value of \( z=4x+2y \) at the corners:
\( \displaystyle (5,0):\ z=20 \)
\( \displaystyle (9,0):\ z=36 \)
\( \displaystyle (0,6):\ z=12 \)
\( \displaystyle (0,5):\ z=10 \)
The region is bounded, so the maximum is the largest corner value, 36, at (9, 0).