The image shows a question related to maximizing and minimizing a linear objective function using a graph of feasible solutions. The graph on the right displays a polygonal region, and you are asked to analyze two objective functions:

(a) $z = 3x + 5y$

(b) $z = x + 6y$

The question focuses on part (b), where you need to find the minimum value of $z = x + 6y$ and determine the points where this minimum occurs.

Steps to solve:

1. Identify the vertices of the feasible region: From the graph, the vertices are:

   • $(2, 10)$
   • $(0, 6)$
   • $(3, 1)$
   • $(7, 3)$

2. Evaluate the objective function $z = x + 6y$ at each vertex:

   • At $(2, 10)$: $z = 2 + 6(10) = 62$
   • At $(0, 6)$: $z = 0 + 6(6) = 36$
   • At $(3, 1)$: $z = 3 + 6(1) = 9$
   • At $(7, 3)$: $z = 7 + 6(3) = 25$

3. Find the minimum value: The minimum value of $z = x + 6y$ occurs at $(3, 1)$ and is $z = 9$.

Conclusion:

• The minimum value is 9, and it occurs at the point $(3, 1)$.

For the answers:

• Choice A is correct: "The minimum value of the objective function $z = x + 6y$ is 9 and is located only at the point $(3, 1)$."

Do you want further details or have any questions?

Here are five related questions:

1. How do you find maximum values using the same method?
2. How can you solve for multiple objective functions at once?
3. What happens if the feasible region is unbounded?
4. How does changing the objective function affect the minimum or maximum values?
5. What are other methods to find extreme values aside from evaluating vertices?

Tip: Always evaluate the objective function at all vertices of the feasible region in linear programming problems.