We are given a linear programming problem with the following constraints and objective function:

Objective Function: $C = -10x + 10y$

Constraints:

1. $y \leq x + 2$
2. $y \leq -3x + 10$
3. $y \geq -x + 4$

Step 1: Graph the Constraints

To determine the feasible region, we need to graph the constraints and find the intersection points. Let's rewrite the constraints in terms of equality to find the boundary lines:

1. $y = x + 2$
2. $y = -3x + 10$
3. $y = -x + 4$

Now, we'll graph these lines and identify the feasible region, which is the area where all constraints are satisfied simultaneously.

Step 2: Find the Intersection Points

We need to find the intersection points of these lines, as the values of $x$ and $y$ that minimize the objective function will occur at the vertices of the feasible region.

Intersection of $y = x + 2$ and $y = -3x + 10$:

Set $x + 2 = -3x + 10$. $x + 2 = -3x + 10$ $x + 3x = 10 - 2$ $4x = 8$ $x = 2$ Substitute $x = 2$ into $y = x + 2$: $y = 2 + 2 = 4$ So, the intersection point is $(2, 4)$.

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

Set $x + 2 = -x + 4$. $x + 2 = -x + 4$ $x + x = 4 - 2$ $2x = 2$ $x = 1$ Substitute $x = 1$ into $y = x + 2$: $y = 1 + 2 = 3$ So, the intersection point is $(1, 3)$.

Intersection of $y = -3x + 10$ and $y = -x + 4$:

Set $-3x + 10 = -x + 4$. $-3x + 10 = -x + 4$ $-3x + x = 4 - 10$ $-2x = -6$ $x = 3$ Substitute $x = 3$ into $y = -x + 4$: $y = -3 + 4 = 1$ So, the intersection point is $(3, 1)$.

Step 3: Evaluate the Objective Function at the Vertices

The vertices of the feasible region are $(2, 4)$, $(1, 3)$, and $(3, 1)$. Now, we evaluate the objective function $C = -10x + 10y$ at each of these points.

1. At $(2, 4)$: $C = -10(2) + 10(4) = -20 + 40 = 20$

2. At $(1, 3)$: $C = -10(1) + 10(3) = -10 + 30 = 20$

3. At $(3, 1)$: $C = -10(3) + 10(1) = -30 + 10 = -20$

Step 4: Minimize the Objective Function

The objective function $C$ achieves its minimum value at the point $(3, 1)$, where $C = -20$.

Conclusion:

The values of $x$ and $y$ that minimize the objective function are $x = 3$ and $y = 1$.

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Would you like more details or have any questions about this process? Here are some additional questions to explore:

1. How can you solve this problem using the Simplex method?
2. What would happen if the constraints were different, such as changing the coefficients of $x$ or $y$?
3. How do we determine whether the solution is a global minimum or a local minimum in linear programming?
4. How can we graph these constraints using a software tool like GeoGebra or Desmos?
5. How would the feasible region change if the inequality signs were reversed?

Tip: When graphing linear programming problems, always check the feasible region's boundaries carefully. The optimal solution often occurs at a vertex, and the objective function should be evaluated at each vertex to find the minimum or maximum.