For the linear programming problem:
Minimize \( z=2x-5y \), subject to the constraints:
\( x+2y\ge 10 \);
\( 3x+4y\le 24 \);
\( x,y\ge 0 \), then optimal value of \( z \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −30
The feasible region lies above x + 2y = 10 and below 3x + 4y = 24, with x, y ≥ 0.
On the y-axis (x = 0): x + 2y ≥ 10 gives y ≥ 5 and 3x + 4y ≤ 24 gives y ≤ 6, so (0, 5) and (0, 6) are corners.
On the x-axis (y = 0) we would need x ≥ 10 and x ≤ 8, which is impossible, so the region does not reach the x-axis.
The two lines meet where
\( \displaystyle x+2y=10,\ 3x+4y=24 \)
\( \displaystyle (3x+4y)-2(x+2y)=24-20 \)
\( \displaystyle x=4,\ y=3 \)
So the feasible region is the triangle with corners (0, 5), (0, 6) and (4, 3). The values of z = 2x − 5y are
\( \displaystyle (0,5):\ z=-25 \)
\( \displaystyle (0,6):\ z=-30 \)
\( \displaystyle (4,3):\ z=8-15=-7 \)
The region is bounded, so the minimum is the smallest corner value: z = −30 at (0, 6).
Why the other options are incorrect
Option (1) −25 — Incorrect. −25 is the value at (0, 5), which is larger than −30.
Option (2) −33 — Incorrect. −33 is not the value of z at any corner of the feasible region.
Option (3) −7 — Incorrect. −7 is the value at (4, 3); it is the maximum of z, not the minimum.