The point at which \(z=3x+4y\) attains maximum value subject to the constraints \(x+y\le 40\), \(x+2y\le 60\), \(x, y\ge 0\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (20, 20)
The maximum of z is reached at a corner point of the feasible region.
The lines x + y = 40 and x + 2y = 60 meet where
\((x+2y)-(x+y)=60-40\)
\(y=20,\quad x=20\)
On the x-axis the tighter limit is x + y ≤ 40, giving (40, 0); on the y-axis it is x + 2y ≤ 60, giving (0, 30). The corners and the values of z = 3x + 4y:
\(z(0, 0) = 0\)
\(z(40, 0) = 120\)
\(z(20, 20) = 60+80 = 140\)
\(z(0, 30) = 120\)
The maximum value 140 is at (20, 20).
Why the other options are incorrect
Option (1) (60, 0) — Incorrect. Here x + y = 60 > 40, so the point is outside the feasible region, even though z = 180 there.
Option (2) (20, 40) — Incorrect. Here x + y = 60 > 40 (and x + 2y = 100 > 60), so the point is not feasible.
Option (4) (40, 0) — Incorrect. It is a corner point, but z = 120 there, less than 140.