Solve the linear programming problem: maximize P = 5x + 3y subject to the constraints x ≥ 0, y ≥ 0, x + y ≤ 10, and 2x + y ≤ 12.
First, we graph the constraints:

The constraint x + y ≤ 10 gives us the line x + y = 10. We shade below this line since we want the sum to be less than or equal to 10.
The constraint 2x + y ≤ 12 gives us the line 2x + y = 12. We shade below this line since we want the sum to be less than or equal to 12.

We are also given that x ≥ 0 and y ≥ 0, so our feasible region is in Quadrant I of the coordinate plane.

Now, we find the vertices of the feasible region:
1. The intersection of x + y = 10 and 2x + y = 12: Solving these equations simultaneously, we get x = 2 and y = 8. So one vertex is (2, 8).
2. The intersection of x + y = 10 and x = 0: This point is (0, 10).
3. The intersection of 2x + y = 12 and x = 0: This point is (0, 12), but it's not part of the feasible region because it doesn't satisfy x + y ≤ 10.
4. The intersection of x + y = 10 and y = 0: This point is (10, 0).
5. The intersection of 2x + y = 12 and y = 0: This point is (6, 0).

So, the vertices of the feasible region are (2, 8), (0, 10), (10, 0), and (6, 0). Now, we evaluate P at each of these points:

P(2, 8) = 5(2) + 3(8) = 10 + 24 = 34
P(0, 10) = 5(0) + 3(10) = 30
P(10, 0) = 5(10) + 3(0) = 50
P(6, 0) = 5(6) + 3(0) = 30

The maximum value of P is 50, which occurs at the point (10, 0).