Minimize $Z = 3x + 5y$
Subject to constraints: $x, y \geq 0, x + 3y – 3 \geq 0, x + y – 2 \geq 0$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 7
The feasible region determined by the system of constraints, $x + 3y \geq 3, x + y \geq 2$, and $x, y \geq 0$ is given below:
Here, the feasible region is unbounded.
The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2)
The values of Z at these corner points are given below:
|
Corner Points |
Coordinates |
Objective Function $Z = 3x + 5y$ |
|
A |
(3, 0) |
9 |
|
B |
(3/2, 1/2) |
7 (Min.) |
|
C |
(0, 2) |
10 |
As we wish to minimize Z, we are going to draw graph of $Z = 3x + 5y < 7$ and check whether the resulting half plane has any common points with the feasibe region or not.
As the inequality, $Z – 3x + 5y < 7$ passes through a corner point B ( 3/2, 1/2) without interfering the feasible region.
That means, the corner point B (3/2, 1/2) minimizes Z and the minimum value of Z is 7. When $x = 3/2, y = 1/2$.