The problem asks you to solve a Linear Programming (LP) problem involving the maximization and minimization of the objective function $p = x + 2y$, subject to the following constraints:

1. $x + y \geq 2$
2. $x + y \leq 6$
3. $x - y \leq 2$
4. $x - y \geq -2$

Steps to Solve:

1. Graph the constraints: Each constraint represents a line, and we need to find the region where all the constraints are satisfied (the feasible region).

   • $x + y \geq 2$: This is a line with slope -1, and the region above it is valid.
   • $x + y \leq 6$: Another line with slope -1, and the region below it is valid.
   • $x - y \leq 2$: This line has a slope of 1, and the region below it is valid.
   • $x - y \geq -2$: This line also has a slope of 1, and the region above it is valid.

2. Identify the vertices: The feasible region is a polygon formed by the intersection of these lines. We calculate the vertices by solving the pairs of line equations at the points where they intersect.

3. Evaluate the objective function at each vertex: The maximum and minimum values of $p = x + 2y$ occur at the vertices of the feasible region.

Finding the vertices:

Vertex 1: Intersection of $x + y = 2$ and $x - y = -2$:

• Solving the system: $x + y = 2$ $x - y = -2$ Add both equations: $2x = 0 \Rightarrow x = 0$ Substitute $x = 0$ into $x + y = 2$: $0 + y = 2 \Rightarrow y = 2$ So, one vertex is $(0, 2)$.

Vertex 2: Intersection of $x + y = 2$ and $x - y = 2$:

• Solving the system: $x + y = 2$ $x - y = 2$ Add both equations: $2x = 4 \Rightarrow x = 2$ Substitute $x = 2$ into $x + y = 2$: $2 + y = 2 \Rightarrow y = 0$ So, another vertex is $(2, 0)$.

Vertex 3: Intersection of $x + y = 6$ and $x - y = -2$:

• Solving the system: $x + y = 6$ $x - y = -2$ Add both equations: $2x = 4 \Rightarrow x = 2$ Substitute $x = 2$ into $x + y = 6$: $2 + y = 6 \Rightarrow y = 4$ So, another vertex is $(2, 4)$.

Vertex 4: Intersection of $x + y = 6$ and $x - y = 2$:

• Solving the system: $x + y = 6$ $x - y = 2$ Add both equations: $2x = 8 \Rightarrow x = 4$ Substitute $x = 4$ into $x + y = 6$: $4 + y = 6 \Rightarrow y = 2$ So, another vertex is $(4, 2)$.

Evaluating the Objective Function:

The vertices are $(0, 2), (2, 0), (2, 4), (4, 2)$.

• At $(0, 2)$: $p = 0 + 2(2) = 4$
• At $(2, 0)$: $p = 2 + 2(0) = 2$
• At $(2, 4)$: $p = 2 + 2(4) = 10$
• At $(4, 2)$: $p = 4 + 2(2) = 8$

Solution:

• Minimum value of $p$: 2, at $(2, 0)$
• Maximum value of $p$: 10, at $(2, 4)$

Would you like further details or explanations on this?

Here are 5 related questions you might explore:

1. What if one of the constraints was changed? How would that affect the feasible region?
2. Can a linear programming problem have more than one optimal solution?
3. How would adding a constraint affect the feasible region?
4. What methods can be used to solve LP problems besides graphing?
5. How does the simplex method work in linear programming?

Tip: Always check if the objective function is unbounded or if the feasible region is empty, as these are potential outcomes in LP problems.