Solve the linear programming problem. Maximize and minimize z equals 2 x plus 3 yz=2x+3y Subject to 2 x plus y2x+y greater than or equals≥ 30 x plus 2 yx+2y greater than or equals≥ 24 x comma yx, y greater than or equals≥ 0.What is the maximum value of​ z? Select the correct choice below and fill in any answer boxes present in your choice.   A. zequals=enter your response here ​(Type an integer or a​ fraction.) B. There is no maximum value of z.

To solve this linear programming problem, we must first identify the feasible region defined by the constraints and then find the value of \( z = 2x + 3y \) at the corner points of the feasible region.

### Problem setup:

- **Objective function**: Maximize (and minimize) \( z = 2x + 3y \)
- **Subject to the constraints**:
  1. \( 2x + y \geq 30 \)
  2. \( x + 2y \geq 24 \)
  3. \( x \geq 0, y \geq 0 \) (non-negativity constraints)

### Step 1: Solve the system of inequalities
To find the corner points, we first solve the system of inequalities by treating them as equalities.

1. **Equation 1**: \( 2x + y = 30 \)
   
   Solve for \( y \) in terms of \( x \):
   \[
   y = 30 - 2x
   \]
   
2. **Equation 2**: \( x + 2y = 24 \)
   
   Solve for \( y \) in terms of \( x \):
   \[
   y = \frac{24 - x}{2}
   \]

### Step 2: Find the intersection points
To find the corner points, we solve the system by finding the intersection of the two lines \( 2x + y = 30 \) and \( x + 2y = 24 \).

Substitute \( y = 30 - 2x \) into \( x + 2y = 24 \):

\[
x + 2(30 - 2x) = 24
\]
\[
x + 60 - 4x = 24
\]
\[
-3x + 60 = 24
\]
\[
-3x = 24 - 60 = -36
\]
\[
x = \frac{-36}{-3} = 12
\]

Now substitute \( x = 12 \) back into the equation \( y = 30 - 2x \):
\[
y = 30 - 2(12) = 30 - 24 = 6
\]

So, one corner point is \( (12, 6) \).

### Step 3: Identify boundary points
We check the boundary points by analyzing the intercepts:
- For \( 2x + y = 30 \):
  - When \( x = 0 \), \( y = 30 \)
  - When \( y = 0 \), \( 2x = 30 \Rightarrow x = 15 \)
  
- For \( x + 2y = 24 \):
  - When \( x = 0 \), \( 2y = 24 \Rightarrow y = 12 \)
  - When \( y = 0 \), \( x = 24 \)

Now, let's summarize the corner points:
- Intersection: \( (12, 6) \)
- Intercepts: \( (15, 0) \), \( (0, 30) \), \( (0, 12) \)

### Step 4: Evaluate \( z = 2x + 3y \) at each corner point
1. At \( (12, 6) \):
   \[
   z = 2(12) + 3(6) = 24 + 18 = 42
   \]

2. At \( (15, 0) \):
   \[
   z = 2(15) + 3(0) = 30
   \]

3. At \( (0, 30) \):
   \[
   z = 2(0) + 3(30) = 90
   \]

4. At \( (0, 12) \):
   \[
   z = 2(0) + 3(12) = 36
   \]

### Step 5: Conclusion
The maximum value of \( z \) occurs at \( (0, 30) \), where \( z = 90 \).

Thus, the correct choice is:
- **A**: \( z = 90 \)

Would you like a detailed explanation of any part of the solution? 

Here are 5 related questions for further practice:
1. How would the solution change if the objective function was \( z = 5x + 2y \)?
2. What is the minimum value of \( z = 2x + 3y \) in the same region?
3. How do we handle linear programming problems with more than two variables?
4. Can you solve for the feasible region graphically for this problem?
5. What happens if one or more constraints are strict inequalities?

**Tip**: Always check the corner points of the feasible region in linear programming to find the optimal solution.

Solve the linear programming problem. Maximize and minimize z = 2x + 3y subject to 2x + y ≥ 30, x + 2y ≥ 24, and x, y ≥ 0. What is the maximum value of z? Select the correct choice below and fill in any answer boxes present in your choice. A. z = (enter your response here) (Type an integer or a fraction.) B. There is no maximum value of z.

Learn how to solve a linear programming problem by maximizing and minimizing the objective function z = 2x + 3y, subject to constraints 2x + y ≥ 30, x + 2y ≥ 24, and non-negativity conditions. This example covers key concepts such as feasible regions and optimization, suitable for high school students in Grades 10-12.

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