Let's analyze the problem step by step.

Objective function:

$z = 50x + 40y$

Vertices of the graphed region:

• A(0, 7)
• B(4, 10)
• C(6, 0)
• D(0, 0)

We will evaluate the objective function $z = 50x + 40y$ at each vertex.

A. Value of the objective function at each corner:

1. At A(0, 7): $z = 50(0) + 40(7) = 0 + 280 = 280$

2. At B(4, 10): $z = 50(4) + 40(10) = 200 + 400 = 600$

3. At C(6, 0): $z = 50(6) + 40(0) = 300 + 0 = 300$

4. At D(0, 0): $z = 50(0) + 40(0) = 0 + 0 = 0$

B. Maximum value of the objective function:

• Among the calculated values: 280, 600, 300, and 0.
• Maximum value = 600 at point B(4, 10).

C. Minimum value of the objective function:

• Minimum value = 0 at point D(0, 0).

Let me know if you need more details or further clarification.

Here are 5 follow-up questions:

1. What is the significance of an objective function in linear programming?
2. How do you determine feasible regions in linear optimization problems?
3. What would change if the coefficients in the objective function were different?
4. Can an objective function have multiple maximum or minimum values?
5. How do constraints affect the shape of the feasible region?

Tip: In linear programming, the optimal solution is always found at a vertex of the feasible region.