Solve the linear programming problem.
Maximize and minimize
z equals 4 x plus 5 y
Subject to 
2 x plus y
greater than or equals
30
x plus 2 y
greater than or equals
24
x comma y
greater than or equals
0
.
.
.
Question content area right
Part 1
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.

### Linear Programming Problem:

You want to:
- Maximize and minimize \( z = 4x + 5y \)

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: Convert the inequalities to equalities (for graphing)
Rearrange the constraints to get:
1. \( 2x + y = 30 \)
2. \( x + 2y = 24 \)

### Step 2: Graph the inequalities
Plot these two lines on the xy-plane and identify the feasible region by finding the points where the inequalities are satisfied. 

### Step 3: Identify corner points
The solution to the linear programming problem will be at one of the vertices (corner points) of the feasible region. The corner points are obtained by:
- Finding the intersections of the lines.
- Using the intercepts with the x- and y-axes.

#### Intersection of the lines:
Solve \( 2x + y = 30 \) and \( x + 2y = 24 \) simultaneously.
From the first equation, solve for \( y \):
\[ y = 30 - 2x \]

Substitute this into the second equation:
\[
x + 2(30 - 2x) = 24
\]
\[
x + 60 - 4x = 24
\]
\[
-3x = -36 \quad \Rightarrow \quad x = 12
\]

Now substitute \( x = 12 \) back into \( y = 30 - 2x \):
\[
y = 30 - 2(12) = 6
\]
Thus, the intersection point is \( (12, 6) \).

#### x-intercept and y-intercept of the lines:
- For \( 2x + y = 30 \):
    - x-intercept: Set \( y = 0 \), then \( x = 15 \).
    - y-intercept: Set \( x = 0 \), then \( y = 30 \).
  
- For \( x + 2y = 24 \):
    - x-intercept: Set \( y = 0 \), then \( x = 24 \).
    - y-intercept: Set \( x = 0 \), then \( y = 12 \).

### Step 4: Test the corner points
The corner points of the feasible region are:
- \( (15, 0) \)
- \( (0, 12) \)
- \( (12, 6) \)

Now, evaluate \( z = 4x + 5y \) at each of these points:

1. At \( (15, 0) \):
   \[
   z = 4(15) + 5(0) = 60
   \]
   
2. At \( (0, 12) \):
   \[
   z = 4(0) + 5(12) = 60
   \]
   
3. At \( (12, 6) \):
   \[
   z = 4(12) + 5(6) = 48 + 30 = 78
   \]

### Step 5: Conclusion
The maximum value of \( z \) is \( 78 \), which occurs at the point \( (12, 6) \).

#### Final Answer:
A. \( z = 78 \)

Would you like further explanation or details on this process? 

---

Here are 5 related questions to expand your understanding:
1. How would the solution change if we wanted to minimize \( z \)?
2. What is the geometric interpretation of the feasible region in linear programming?
3. How do slack variables work in converting inequalities into equalities for linear programming?
4. Can you solve this problem graphically by sketching the region of constraints?
5. How does the Simplex Method approach linear programming compared to the graphical method?

Tip: In linear programming, the optimal solution always lies at a corner (vertex) of the feasible region.

Solve the linear programming problem. Maximize and minimize z = 4x + 5y subject to the constraints 2x + y ≥ 30, x + 2y ≥ 24, and x, y ≥ 0. What is the maximum value of z?

Learn how to solve a linear programming problem to maximize z = 4x + 5y. This solution involves finding the intersection of constraints and applying the corner point theorem to identify the maximum value. Suitable for students in Grades 10-12.

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