Consider a linear programming problem with the following constraints and objective function, C.

y≤x+2

y≤–3x+10

y≥–x+4

C=–10x+10y

What are the values of x and y that minimize the objective function for the feasible region?

To minimize the objective function \( C = -10x + 10y \) subject to the constraints, we will:

1. **Graph the Constraints**: Plot the three inequalities to determine the feasible region.
2. **Find the Corner Points**: Identify the coordinates of the corner points of the feasible region.
3. **Evaluate the Objective Function at Each Corner**: Calculate \( C = -10x + 10y \) at each of the corner points to find the minimum value.

### Step 1: Graph the Constraints

The constraints are:
- \( y \leq x + 2 \)
- \( y \leq -3x + 10 \)
- \( y \geq -x + 4 \)

These describe half-planes, and the feasible region is the intersection of all these half-planes. We will first express the equations of the lines:
1. \( y = x + 2 \)
2. \( y = -3x + 10 \)
3. \( y = -x + 4 \)

These are straight lines, and we need to determine which side of each line satisfies the inequality. 

### Step 2: Find the Intersection Points (Corner Points)

We can find the intersections by solving pairs of equations:

1. **Intersection of \( y = x + 2 \) and \( y = -3x + 10 \)**:
   \[
   x + 2 = -3x + 10 \implies 4x = 8 \implies x = 2
   \]
   Substituting \( x = 2 \) into \( y = x + 2 \):
   \[
   y = 2 + 2 = 4
   \]
   So, the intersection point is \( (2, 4) \).

2. **Intersection of \( y = x + 2 \) and \( y = -x + 4 \)**:
   \[
   x + 2 = -x + 4 \implies 2x = 2 \implies x = 1
   \]
   Substituting \( x = 1 \) into \( y = x + 2 \):
   \[
   y = 1 + 2 = 3
   \]
   So, the intersection point is \( (1, 3) \).

3. **Intersection of \( y = -3x + 10 \) and \( y = -x + 4 \)**:
   \[
   -3x + 10 = -x + 4 \implies -2x = -6 \implies x = 3
   \]
   Substituting \( x = 3 \) into \( y = -x + 4 \):
   \[
   y = -3 + 4 = 1
   \]
   So, the intersection point is \( (3, 1) \).

### Step 3: Evaluate the Objective Function at Each Corner

Now we evaluate the objective function \( C = -10x + 10y \) at each of the corner points \( (2, 4) \), \( (1, 3) \), and \( (3, 1) \):

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: Conclusion

The objective function \( C = -10x + 10y \) achieves its minimum value at the point \( (3, 1) \), where \( C = -20 \).

### Final Answer:
The values of \( x \) and \( y \) that minimize the objective function are \( \boxed{x = 3} \) and \( \boxed{y = 1} \).

---

Would you like further details on any part of the process? 

Here are some related questions to explore:

1. How do you graph a system of inequalities to find the feasible region?
2. What are corner points in a linear programming problem and why are they important?
3. How does the simplex method differ from graphical methods in linear programming?
4. Can you solve a linear programming problem without graphing?
5. What happens if the objective function is maximized instead of minimized? 

**Tip**: Always ensure that the intersection points are within the feasible region defined by the inequalities before evaluating the objective function.

Consider a linear programming problem with the following constraints and objective function, C.

y ≤ x + 2

y ≤ -3x + 10

y ≥ -x + 4

C = -10x + 10y

What are the values of x and y that minimize the objective function for the feasible region?

Learn how to minimize the objective function C = -10x + 10y subject to given constraints using linear programming. This problem involves graphing inequalities, identifying feasible regions, and evaluating corner points to find the optimal solution. Suitable for students in Grades 10-12.

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