Solve the following linear programming problem.
​Maximize:
zequals21xplus11y
subject​ to:
3xplus5yless than or equals15
7xplusyless than or equals15
xgreater than or equals​0, ygreater than or equals0
Question content area bottom left
Part 1
The maximum value is

15.
Part 2
The maximum occurs at the point

enter your response here.

The given linear programming problem is:

**Maximize:**
\[ z = 21x + 11y \]

**Subject to:**
1. \( 3x + 5y \leq 15 \)
2. \( 7x + y \leq 15 \)
3. \( x \geq 0, \, y \geq 0 \)

### Step 1: Graph the constraints
We will first convert the inequalities to equalities and find the boundary points.

1. For \( 3x + 5y = 15 \):
   - When \( x = 0 \), \( y = 3 \).
   - When \( y = 0 \), \( x = 5 \).
   Thus, this line passes through \( (5, 0) \) and \( (0, 3) \).

2. For \( 7x + y = 15 \):
   - When \( x = 0 \), \( y = 15 \).
   - When \( y = 0 \), \( x = \frac{15}{7} \approx 2.14 \).
   Thus, this line passes through \( (2.14, 0) \) and \( (0, 15) \).

### Step 2: Find the feasible region
Plotting the two lines on a graph, along with the conditions \( x \geq 0 \) and \( y \geq 0 \), the feasible region is the area where all constraints are satisfied. The vertices of this region are the points where the boundary lines intersect or hit the axes.

### Step 3: Identify the corner points
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (5, 0) \)
3. \( (0, 3) \)
4. The intersection of the lines \( 3x + 5y = 15 \) and \( 7x + y = 15 \).

#### Finding the intersection of the lines:
Solve the system of equations:

1. \( 3x + 5y = 15 \)
2. \( 7x + y = 15 \)

Multiply the second equation by 5 to eliminate \( y \):

\[
7x + y = 15 \quad \Rightarrow \quad 35x + 5y = 75
\]

Subtract the first equation from this:

\[
(35x + 5y) - (3x + 5y) = 75 - 15
\]
\[
32x = 60 \quad \Rightarrow \quad x = \frac{60}{32} = 1.875
\]

Substitute \( x = 1.875 \) into \( 7x + y = 15 \):

\[
7(1.875) + y = 15 \quad \Rightarrow \quad 13.125 + y = 15 \quad \Rightarrow \quad y = 1.875
\]

Thus, the intersection point is \( (1.875, 1.875) \).

### Step 4: Evaluate the objective function
Evaluate \( z = 21x + 11y \) at each corner point:

1. At \( (0, 0) \): 
   \[
   z = 21(0) + 11(0) = 0
   \]
2. At \( (5, 0) \):
   \[
   z = 21(5) + 11(0) = 105
   \]
3. At \( (0, 3) \):
   \[
   z = 21(0) + 11(3) = 33
   \]
4. At \( (1.875, 1.875) \):
   \[
   z = 21(1.875) + 11(1.875) = 39.375 + 20.625 = 60
   \]

### Step 5: Conclusion
The maximum value of \( z = 105 \) occurs at the point \( (5, 0) \).

---

Would you like more details on any of the steps?

Here are 5 follow-up questions that might be of interest:
1. How do we interpret the feasible region in linear programming?
2. Can you explain how the graphical method works for linear programming problems?
3. What happens when the maximum point lies on the intersection of two boundary lines?
4. How can we use the simplex method to solve this problem?
5. What is the difference between linear and nonlinear programming?

**Tip:** In linear programming, the maximum or minimum value of the objective function always occurs at one of the vertices (corner points) of the feasible region.

Solve the following linear programming problem. Maximize: z = 21x + 11y subject to: 3x + 5y ≤ 15, 7x + y ≤ 15, x ≥ 0, y ≥ 0

Solve a linear programming maximization problem with objective function z = 21x + 11y and constraints 3x + 5y ≤ 15, 7x + y ≤ 15, x ≥ 0, y ≥ 0. This detailed solution walks through graphing the constraints, identifying corner points, and applying the corner point theorem. Suitable for Grades 10-12.

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