Question

What are the two trivial constraints?

Answer 1

The trivial constraints represent the boundaries of the graph within the first quadrant.
- As the graph is confined to the first quadrant, the two trivial constraints are \( x \geq 0 \) and \( y \geq 0 \).

Answer 2

Step 1: One non-trivial constraint is \( x + 2y \geq 100 \). The graph indicates the following lines:
- The line through \( A(0, 50) \) and \( E(100, 0) \) represents \( x + 2y = 100 \).
- The line through \( B(20, 40) \) and \( C(50, 100) \) represents \( y - \frac{4}{3}x + \frac{80}{3} \geq 0 \).
- The line through \( C(50, 100) \) and \( D(0, 200) \) represents \( y - 2x \leq 0 \).
Final Answer: The remaining two non-trivial constraints are: \[ y - \frac{4}{3}x + \frac{80}{3} \geq 0 \quad {and} \quad y - 2x \leq 0. \]

Answer 3

Region \( R_2 \) is defined by \( x + 2y \geq 100 \), \( x \geq 0 \), and \( y \geq 0 \), excluding the constraints of region \( R_1 \). The constraints, derived from the graph, are \( x + 2y \geq 100 \) and \( y - \frac{4}{3}x + \frac{80}{3} \leq 0 \).
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 to be maximized within the feasible region \( R_1 \). The vertices defining \( R_1 \) are \( A(0, 50) \), \( B(20, 40) \), and \( C(50, 100) \).
Step 2: Calculate the value of \( z \) at each vertex: - For \( A(0, 50) \): \( z = 5(0) + 2(50) = 100 \). - For \( B(20, 40) \): \( z = 5(20) + 2(40) = 180 \). - For \( C(50, 100) \): \( z = 5(50) + 2(100) = 450 \). 
Step 3: The maximum value of \( z = 5x + 2y \) is 450, occurring at vertex \( C(50, 100) \).