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

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 2

What is the maximum profit?

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 \( \boxed{14} \).