Question

Graph the system of constraints and find the values of x and y that maximize P for the objective function P = 3x + 2y x+y ≤ 6 2x+y ≤ 8 x ≥ 0 Y ≥ 0 What are the values of x and y that maximize P?

Answer 1

What is the maximum profit?

Answer 2

To maximize the objective function \( P = 3x + 2y \) subject to the constraints:

1. \( x + y \leq 6 \)
2. \( 2x + y \leq 8 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

### Step 1: Graph the Constraints

1. **Graph \( x + y \leq 6 \)**:
- The line \( x + y = 6 \) has intercepts at \( (6, 0) \) and \( (0, 6) \).
- Below this line represents the region for the inequality.

2. **Graph \( 2x + y \leq 8 \)**:
- The line \( 2x + y = 8 \) intersects the axes at \( (4, 0) \) and \( (0, 8) \).
- The region below this line represents the inequality.

3. **Non-negativity Constraints**:
- \( x \geq 0 \) means we are only considering the right half of the plane.
- \( y \geq 0 \) means we are only considering the top half of the plane.

### Step 2: Find the Feasible Region

To find the feasible region, we will graph the lines and see where they intersect:

- The intersection of \( x + y = 6 \) and \( 2x + y = 8 \):
\[
\begin{align*}
x + y &= 6 \quad \text{(1)} \\
2x + y &= 8 \quad \text{(2)}
\end{align*}
\]
Subtracting (1) from (2):
\[
2x + y - (x + y) = 8 - 6 \\
x = 2
\]
Substitute \( x = 2 \) into equation (1):
\[
2 + y = 6 \implies y = 4
\]
So, the intersection point is \( (2, 4) \).

Next, we find the vertices of the feasible region, which are determined by the intersection points and the axes:
- The line \( x + y = 6 \) intersects the axes at \( (6, 0) \) and \( (0, 6) \).
- The line \( 2x + y = 8 \) intersects the axes at \( (4, 0) \) and \( (0, 8) \).

The intersecting points inside the non-negative quadrant that satisfy both inequalities are:
1. \( (0, 0) \) (origin)
2. \( (0, 6) \) (from \( x+y=6 \))
3. \( (4, 0) \) (from \( 2x+y=8 \))
4. \( (2, 4) \) (intersection point)

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

Now we evaluate \( P = 3x + 2y \) at the vertices of the feasible region:
1. At \( (0, 0) \):
\[
P = 3(0) + 2(0) = 0
\]
2. At \( (0, 6) \):
\[
P = 3(0) + 2(6) = 12
\]
3. At \( (4, 0) \):
\[
P = 3(4) + 2(0) = 12
\]
4. At \( (2, 4) \):
\[
P = 3(2) + 2(4) = 6 + 8 = 14
\]

### Step 4: Find the Maximum Value

From our evaluations, we have:
- \( P(0, 0) = 0 \)
- \( P(0, 6) = 12 \)
- \( P(4, 0) = 12 \)
- \( P(2, 4) = 14 \)

The maximum value is \( 14 \), which occurs at the point \( (2, 4) \).

### Conclusion

The values of \( x \) and \( y \) that maximize \( P \) are \( \boxed{(2, 4)} \).

Answer 3

The maximum profit, represented by the objective function \( P = 3x + 2y \), occurs at the point \( (2, 4) \). At this point, we previously calculated the profit as follows:

\[
P = 3(2) + 2(4) = 6 + 8 = 14.
\]

Thus, the maximum profit is \( \boxed{14} \).