Question

A company produces mopeds and bicycles. It must produce at least 10 mopeds per month. The company has the equipment to produce only 60 mopeds. It also can produce only 120 bicycles. The production of mopeds and bicycles cannot exceed 160. The profit on a moped is $134, and on a bicycle, it's $20. How many of each should be manufactured per month to maximize profit?

A. P = 20x + 134y
B. P = 134x + 20y
C. P = 60x + 120y
D. P = 120x + 60y

Answer 1

The company should produce mopeds and bicycles in quantities that maximize the profit function P = 134x + 20y while adhering to the constraints of producing at least 10 and no more than 60 mopeds, no more than 120 bicycles, and the total production not exceeding 160 units. Correct option is B) P = 134x + 20y

Step-by-step explanation:

To solve this problem, we need to determine the combination of mopeds and bicycles the company should produce to maximize profit under the given constraints. We are given two constraints aside from non-negativity: the company must produce at least 10 mopeds and cannot produce more than 60 mopeds, and it cannot produce more than 120 bicycles. Also, the total production of mopeds and bicycles cannot exceed 160.

Let's denote the number of mopeds by x and the number of bicycles by y. The profit function based on the question is P = 134x + 20y. The constraints are:

• x ≥ 10 (at least 10 mopeds)
• x ≤ 60 (no more than 60 mopeds)
• y ≤ 120 (no more than 120 bicycles)
• x + y ≤ 160 (total production not to exceed 160 units)

To maximize profit, we need to find the values of x and y that satisfy these constraints and yield the highest value for the profit function P = 134x + 20y. This is a linear optimization problem that can be solved using methods such as graphical representation or linear programming (such as the Simplex algorithm).