To maximize the objective function Z = 3x + 4y, we need to find the maximum value of Z subject to the given constraints.

Constraint 1: x + y ≤ 4
This constraint represents the restriction that the sum of x and y should be less than or equal to 4.

Constraint 2: x ≥ 0
This constraint represents the restriction that x should be greater than or equal to 0.

Constraint 3: y ≥ 0
This constraint represents the restriction that y should be greater than or equal to 0.

To find the maximum value of Z, we need to consider the feasible region defined by the intersection of the three constraints.

Solving the constraints:
1. For the first constraint, x + y ≤ 4, we can plot the line x + y = 4 on a graph. This line passes through the points (0, 4) and (4, 0) and divides the graph into two regions. We are interested in the region where x + y is less than or equal to 4.

2. For the second constraint, x ≥ 0, we know that x should be greater than or equal to 0. This constraint limits the feasible region to the positive x-axis and the right half of the graph.

3. For the third constraint, y ≥ 0, we know that y should be greater than or equal to 0. This constraint limits the feasible region to the positive y-axis and the upper half of the graph.

The feasible region is the intersection of the three constraints, which is the triangle formed by the points (0, 4), (4, 0), and (0, 0).

To find the maximum value of Z, we evaluate the objective function Z = 3x + 4y at the corner points of the feasible region.

Corner points of the feasible region:
1. (0, 0)
2. (0, 4)
3. (4, 0)

Evaluate Z at the corner points:
1. Z = 3(0) + 4(0) = 0
2. Z = 3(0) + 4(4) = 16
3. Z = 3(4) + 4(0) = 12

The maximum value of Z is 16, which occurs at the point (0, 4).

Therefore, the correct answer is option 'C': Maximum Z = 16 at (0, 4).