For the LP: Max Z = X1 + X2, subject to X1 + X2 ≤ 4, 2X1 + 5X2 ≤ 10, X1 ≥ 0, X2 ≥ 0, which of the following is the optimal point (X1, X2) and its Z value?

• A. (3,0), Z=3
• B. (1,1), Z=2
• C. (4,0), Z=4
• D. (2,1), Z=3

Answer: (C)

Maximizing Z = X1 + X2 means moving to the northeast within the feasible region, pushing for the largest possible sum X1 + X2. The constraint X1 + X2 ≤ 4 caps the sum at 4, so the best you can do is Z = 4 if that value is feasible with the other constraint.

Check a point on that boundary that also satisfies the second constraint: (4,0) gives Z = 4, and 2X1 + 5X2 = 8 ≤ 10, so it is feasible. Therefore the maximum value of Z is 4, and (4,0) is an optimal point. In fact, the line X1 + X2 = 4 intersects 2X1 + 5X2 = 10 at (10/3, 2/3), so every point on the segment between (10/3, 2/3) and (4,0) yields Z = 4, i.e., multiple optimal solutions exist.

The other provided points produce smaller objective values (3 or 2), so they are not optimal.