Step 1: Graph the constraints

First, we convert the inequalities into equations to graph the lines:

• 3x + 2y = 9
• 3x + y = 9
• x = 0
• y = 0

For 3x + 2y = 9:

• When x = 0, y = 4.5
• When y = 0, x = 3

For 3x + y = 9:

• When x = 0, y = 9
• When y = 0, x = 3

Graph these lines on the x-y plane. The feasible region is bounded by these lines and the axes (x ≥ 0, y ≥ 0).

Step 2: Identify the feasible region

The feasible region is the area that satisfies all the inequalities. In this case, it is the region bounded by the lines 3x + 2y ≤ 9, 3x + y ≤ 9, x ≥ 0, and y ≥ 0.

Step 3: Find the corner points of the feasible region

The corner points are the vertices of the feasible region. From the graph, we can identify the following corner points:

• (0, 0)
• (3, 0)
• (0, 4.5)

We also need to find the intersection of the two lines 3x + 2y = 9 and 3x + y = 9. Subtracting the second equation from the first gives:

y = 0

Substituting y = 0 into 3x + y = 9 gives:

3x = 9

x = 3

So, the intersection point is (3, 0). However, this point is already listed.

To find the intersection of 3x + 2y = 9 and 3x + y = 9, we can solve the system of equations:

3x + 2y = 9

3x + y = 9

Subtracting the second equation from the first gives y = 0. Substituting y = 0 into the second equation gives 3x = 9, so x = 3. The intersection point is (3, 0).

Let's find the intersection of 3x + 2y = 9 and x = 0. When x = 0, 2y = 9, so y = 4.5. The point is (0, 4.5).

Let's find the intersection of 3x + y = 9 and x = 0. When x = 0, y = 9. The point is (0, 9).

The intersection of y = 0 with 3x + 2y = 9 gives 3x = 9, so x = 3. The point is (3, 0).

The intersection of y = 0 with 3x + y = 9 gives 3x = 9, so x = 3. The point is (3, 0).

To find the intersection of 3x + 2y = 9 and 3x + y = 9, subtract the second equation from the first: y = 0. Then 3x + 0 = 9, so x = 3. The point is (3, 0).

The corner points are (0, 0), (3, 0), and (0, 4.5). We need to find the intersection of 3x+2y=9 and 3x+y=9. Subtracting the equations gives y=0. Substituting into the second equation gives 3x=9, so x=3. Thus, the intersection is (3,0).

However, the feasible region is also bounded by the intersection of 3x+2y=9 and 3x+y=9. Subtracting the second from the first gives y=0. Substituting into the second gives 3x=9, so x=3. The point is (3,0).

The correct corner points are (0,0), (3,0), and (0, 4.5).

Step 4: Evaluate the objective function at each corner point

P = 70x + 40y

• At (0, 0): P = 70(0) + 40(0) = 0
• At (3, 0): P = 70(3) + 40(0) = 210
• At (0, 4.5): P = 70(0) + 40(4.5) = 180

Step 5: Determine the maximum value

The maximum value of P is 210, which occurs at the point (3, 0).