The maximum value of \(z=2.5x+y\) subject to the constraints \(x+3y\le 12\), \(3x+y\le 12\), \(x, y\ge 0\), is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10.5
The feasible region is bounded by the axes and the two lines. The lines meet where
\(x + 3y = 12\)
\(3x + y = 12\)
Subtracting gives 2x − 2y = 0, so x = y, and then 4x = 12:
\(x = 3,\quad y = 3\)
The corner points are (0, 0), (4, 0), (3, 3) and (0, 4). Values of z = 2.5x + y:
\(z(0, 0) = 0\)
\(z(4, 0) = 10\)
\(z(3, 3) = 7.5 + 3 = 10.5\)
\(z(0, 4) = 4\)
So the maximum value of z is 10.5, at (3, 3).
Why the other options are incorrect
Option (1) 4 — Incorrect. This is z at (0, 4), not the largest corner value.
Option (2) 8.5 — Incorrect. No corner gives 8.5; the largest value is 10.5.
Option (4) 10 — Incorrect. This is z at (4, 0); the corner (3, 3) gives the larger value 10.5.