The maximum value of \( z=22x+18y \) subject to the constraints:
\( x+y\le 20, \)
\( 3x+2y\le 48,\ x,y\ge 0 \) is attained at the point:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (8, 12)
The lines x + y = 20 and 3x + 2y = 48 meet the axes at (20, 0), (0, 20) and (16, 0), (0, 24). In the first quadrant the region is bounded by (0, 20) on the y-axis and (16, 0) on the x-axis.
The two lines meet where
\( \displaystyle x+y=20 \)
\( \displaystyle 3x+2y=48 \)
Subtracting twice the first from the second gives x = 8, and then y = 12.
So the corner points of the feasible region are (0, 0), (16, 0), (8, 12) and (0, 20).
\( \displaystyle z(0,0)=0 \)
\( \displaystyle z(16,0)=352 \)
\( \displaystyle z(8,12)=176+216=392 \)
\( \displaystyle z(0,20)=360 \)
The maximum 392 is attained at (8, 12).