Question

What are the two trivial constraints?

Answer 1

Trivial constraints correspond to the boundaries of the graph in the first quadrant. 
- Since the graph is in the first quadrant, \( x \geq 0 \) and \( y \geq 0 \) are the two trivial constraints.

Answer 2

Step 1: Given one non-trivial constraint is \( x + 2y \geq 100 \). From the graph: 
- The line passing through \( A(0, 50) \) and \( E(100, 0) \) corresponds to \( x + 2y = 100 \). 
- The line passing through \( B(20, 40) \) and \( C(50, 100) \) corresponds to \( y - \frac{4}{3}x + \frac{80}{3} \geq 0 \). 
- The line passing through \( C(50, 100) \) and \( D(0, 200) \) corresponds to \( y - 2x \leq 0 \). 
Final Answer: The other two non-trivial constraints are: \[ y - \frac{4}{3}x + \frac{80}{3} \geq 0 \quad {and} \quad y - 2x \leq 0. \]

Answer 3

For \( R_2 \), the region includes \( x + 2y \geq 100 \), \( x \geq 0 \), and \( y \geq 0 \), but excludes the constraints forming the region \( R_1 \). Based on the graph, the constraints are derived as: - \( x + 2y \geq 100 \), - \( y - \frac{4}{3}x + \frac{80}{3} \leq 0 \). 
Final Answer: The constraints for \( R_2 \) are: \[ x + 2y \geq 100 \quad {and} \quad y - \frac{4}{3}x + \frac{80}{3} \leq 0. \]

Answer 4

Step 1: The objective function \( z = 5x + 2y \) is maximized at one of the vertices of the feasible region \( R_1 \). The vertices of \( R_1 \) are \( A(0, 50) \), \( B(20, 40) \), and \( C(50, 100) \). 
Step 2: Evaluate \( z \) at each vertex: - At \( A(0, 50) \): \[ z = 5(0) + 2(50) = 100. \] - At \( B(20, 40) \): \[ z = 5(20) + 2(40) = 100 + 80 = 180. \] - At \( C(50, 100) \): \[ z = 5(50) + 2(100) = 250 + 200 = 450. \] 
Step 3: Conclusion: The maximum value of \( z = 5x + 2y \) is 450 at \( C(50, 100) \).