The problem involves an objective function $z = 30x + 50y$, and the task is to evaluate this function at various corner points of the graphed region. Let's solve the questions step by step:

Given Points:

• A: (0, 6)
• B: (3, 10)
• C: (6, 0)
• D: (0, 0)

Part A: Find the value of the objective function at each corner.

The formula is $z = 30x + 50y$, so we will plug in the coordinates of each point.

• At Point A (0, 6): $z = 30(0) + 50(6) = 0 + 300 = 300$ So, the value at point A is 300 (already given).

• At Point B (3, 10): $z = 30(3) + 50(10) = 90 + 500 = 590$ The value at point B is 590.

• At Point C (6, 0): $z = 30(6) + 50(0) = 180 + 0 = 180$ The value at point C is 180.

• At Point D (0, 0): $z = 30(0) + 50(0) = 0$ The value at point D is 0.

Part B: Maximum value of the objective function

From the calculated values, the maximum value is 590 at point B (3, 10).

Part C: Minimum value of the objective function

The minimum value is 0 at point D (0, 0).

Do you want more details on how this was done, or do you have any further questions?

Here are five related questions for deeper understanding:

1. How would the solution change if the objective function had different coefficients, such as $z = 20x + 40y$?
2. How do you determine which points to evaluate in a linear programming problem?
3. Can the minimum and maximum values be equal in certain linear programming problems?
4. What if there are constraints that further limit the feasible region?
5. How would you solve this graphically if more constraints were added?

Tip: Always evaluate the objective function at the corner points (vertices) of the feasible region in linear programming problems to find the optimal values.