what are the maximum points for the highest value of z

Let's calculate the maximum value of \( z = 10x + 10y \) by first finding the corner points of the feasible region formed by the constraints. After that, we will evaluate the value of \( z \) at each of these points.

### Constraints Recap:
1. \( 8x + 6y \geq 120 \)
2. \( 14x - 11y \geq 38 \)
3. \( x + y \leq 42 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 1: Find the Intersection Points
We'll find the points of intersection for the pairs of constraint lines to determine the vertices of the feasible region.

#### 1. Intersection of \( 8x + 6y = 120 \) and \( 14x - 11y = 38 \):
- Solve these two equations simultaneously.

\[
\text{Equation 1: } 8x + 6y = 120
\]
\[
\text{Equation 2: } 14x - 11y = 38
\]

Multiply Equation 1 by 11 and Equation 2 by 6 to eliminate \( y \):

\[
88x + 66y = 1320
\]
\[
84x - 66y = 228
\]

Add these equations:

\[
172x = 1548
\]
\[
x = 9
\]

Substitute \( x = 9 \) back into Equation 1:

\[
8(9) + 6y = 120 \rightarrow 72 + 6y = 120 \rightarrow 6y = 48 \rightarrow y = 8
\]

So, the intersection point is \( (9, 8) \).

#### 2. Intersection of \( 8x + 6y = 120 \) and \( x + y = 42 \):
- Substitute \( y = 42 - x \) into \( 8x + 6y = 120 \):

\[
8x + 6(42 - x) = 120
\]
\[
8x + 252 - 6x = 120 \rightarrow 2x = -132 \rightarrow x = -66
\]

Since \( x = -66 \) is not within the feasible region \( x \geq 0 \), this intersection does not contribute to a valid vertex in the feasible region.

#### 3. Intersection of \( 14x - 11y = 38 \) and \( x + y = 42 \):
- Substitute \( y = 42 - x \) into \( 14x - 11y = 38 \):

\[
14x - 11(42 - x) = 38
\]
\[
14x - 462 + 11x = 38 \rightarrow 25x = 500 \rightarrow x = 20
\]

Substitute \( x = 20 \) into \( y = 42 - x \):

\[
y = 42 - 20 = 22
\]

So, the intersection point is \( (20, 22) \).

#### 4. Intersection of \( 8x + 6y = 120 \) and \( y = 0 \) (substituting \( y = 0 \)):
- Substitute \( y = 0 \) into \( 8x + 6y = 120 \):

\[
8x + 0 = 120 \rightarrow x = 15
\]

So, the intersection point is \( (15, 0) \).

#### 5. Intersection of \( x = 0 \) and \( 8x + 6y = 120 \):
- Substitute \( x = 0 \) into \( 8x + 6y = 120 \):

\[
6y = 120 \rightarrow y = 20
\]

So, the intersection point is \( (0, 20) \).

### Step 2: Evaluate \( z = 10x + 10y \) at Each Vertex
The vertices of the feasible region are:
1. \( (9, 8) \)
2. \( (20, 22) \)
3. \( (15, 0) \)
4. \( (0, 20) \)

Now, evaluate \( z \) at each vertex:
- \( z(9, 8) = 10(9) + 10(8) = 90 + 80 = 170 \)
- \( z(20, 22) = 10(20) + 10(22) = 200 + 220 = 420 \)
- \( z(15, 0) = 10(15) + 10(0) = 150 + 0 = 150 \)
- \( z(0, 20) = 10(0) + 10(20) = 0 + 200 = 200 \)

### Conclusion: 
The maximum value of \( z = 420 \) occurs at the point \( (20, 22) \).

Would you like any additional details or explanations?

---

### Follow-up Questions:
1. Would you like to see the feasible region graphically?
2. Are you interested in verifying these calculations for the other constraints?
3. Do you need further clarification on how the vertices were determined?
4. Should we explore how changes in constraints affect the maximum \( z \) value?
5. Would you like to learn more about different methods for solving linear programming problems?

### Tip:
In linear programming, the maximum or minimum value of the objective function always occurs at one of the vertices of the feasible region. Checking each vertex systematically ensures you find the optimal solution.

Learn how to maximize the objective function z = 10x + 10y using linear programming techniques. This problem involves finding intersection points of constraints and evaluating z at these points to determine the maximum value. Suitable for advanced high school students or anyone interested in linear programming.

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