To solve this problem step by step, we aim to find the minimum value of $z = x - 3y$ at the vertices of the feasible region shown in the graph. Here's how to solve:

Step 1: Identify the vertices

The feasible region is a polygon whose vertices are clearly labeled in the graph. The vertices are:

• $(1, 1)$
• $(3, 8)$
• $(6, 11)$
• $(7, 4)$

Step 2: Compute $z = x - 3y$ at each vertex

Substitute the coordinates of each vertex into the expression $z = x - 3y$:

1. At $(1, 1)$: $z = 1 - 3(1) = 1 - 3 = -2$

2. At $(3, 8)$: $z = 3 - 3(8) = 3 - 24 = -21$

3. At $(6, 11)$: $z = 6 - 3(11) = 6 - 33 = -27$

4. At $(7, 4)$: $z = 7 - 3(4) = 7 - 12 = -5$

Step 3: Determine the minimum value

The calculated $z$-values are:

• $z = -2$ at $(1, 1)$
• $z = -21$ at $(3, 8)$
• $z = -27$ at $(6, 11)$
• $z = -5$ at $(7, 4)$

The minimum value is: $z = -27 \text{ at } (6, 11)$

Final Answer:

The minimum value of $z = x - 3y$ is -27, and it occurs at the vertex $(6, 11)$.

Do you want a detailed explanation of any step? Here are related questions to deepen your understanding:

1. How do we find feasible regions in optimization problems?
2. Why do we evaluate $z$ only at the vertices of the region?
3. What happens if $z$ has multiple minima?
4. How can we solve linear programming problems algebraically?
5. What are practical applications of minimizing or maximizing such functions?

Tip: Always verify your feasible region before evaluating $z$; small graphing errors can lead to incorrect results!