A factory makes two products, X and Y. The profit per unit is ¥40,000 for X and ¥30,000 for Y; making one unit takes 2 hours of machine time for X and 1 hour for Y, and 1 hour of worker time for both X and Y. If up to 100 machine hours and 80 worker hours are available per week, which is the maximum weekly profit?

Answer: B. ¥2.6 million

If the quantities of X and Y are x and y, the constraints are 2x + y ≤ 100 (machine) and x + y ≤ 80 (workers), and maximizing the profit 4x + 3y (in ¥10,000) is a linear programming problem. The maximum is reached at a vertex of the region bounded by the constraints; solving for the intersection of the two constraint lines gives x = 20 and y = 60, for a profit of 4 × 20 + 3 × 60 = 260, that is, ¥2.6 million. At the other vertices, making only 80 units of Y gives ¥2.4 million and making only 50 units of X gives ¥2 million, so making a combination of X and Y is the best.

• A. This is the profit from making only 80 units of Y. Machine time is left over, so combining in some X increases profit.
• B. Correct. The maximum is at x = 20, y = 60, which uses exactly all the machine and worker hours.
• C. This ignores the worker-hour constraint and makes 100 units of Y using machine time alone. That would need more than 80 worker hours, so it is not feasible.