Question

For the given graph of a Linear Programming Problem, write all the constraints satisfying the given feasible region.

Hint:

To write the constraints for a Linear Programming Problem from a graph, find the equations of the lines forming the boundaries of the feasible region and translate them into inequalities.

Answer 1

From the given graph, we can identify the constraints that form the feasible region. The vertices of the region are $ A(0, 200) $, $ B(50, 250) $, $ C(150, 150) $, and $ D(200, 0) $. Using the equation of the line passing through any two points, we can write the inequalities corresponding to the constraints. 1. From Point A(0, 200) to B(50, 250):
The slope \( m \) is calculated as: \[ m = \frac{250 - 200}{50 - 0} = \frac{50}{50} = 1 \] Using the point \( A(0, 200) \) in the equation of the line: \[ y - 200 = 1(x - 0) \quad \Rightarrow \quad y = x + 200 \] Thus, the first constraint is: \[ y \leq x + 200 \] 2. From Point B(50, 250) to C(150, 150):
The slope \( m \) is: \[ m = \frac{150 - 250}{150 - 50} = \frac{-100}{100} = -1 \] Using point \( B(50, 250) \): \[ y - 250 = -1(x - 50) \quad \Rightarrow \quad y = -x + 300 \] Thus, the second constraint is: \[ y \leq -x + 300 \] 3. From Point C(150, 150) to D(200, 0):
The slope \( m \) is: \[ m = \frac{0 - 150}{200 - 150} = \frac{-150}{50} = -3 \] Using point \( C(150, 150) \): \[ y - 150 = -3(x - 150) \quad \Rightarrow \quad y = -3x + 750 \] Thus, the third constraint is: \[ y \leq -3x + 750 \] 4. From Point D(200, 0) to O(0, 0):
The slope \( m \) is: \[ m = \frac{0 - 0}{200 - 0} = 0 \] The equation is simply: \[ y = 0 \] Thus, the fourth constraint is: \[ y \geq 0 \] Constraints:
The constraints for the feasible region are: \[ y \leq x + 200 \] \[ y \leq -x + 300 \] \[ y \leq -3x + 750 \] \[ y \geq 0 \] These constraints define the feasible region in the given Linear Programming Problem.