Question

2- A factory manufactures two types of gadgets, regular and premium. Each gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing, while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other restrictions, the company can make at most 7 gadgets a day. If a profit of $20 is realized for each regular gadget and $30 for a premium gadget, how many of each should be manufactured to maximize profit?

Answer 1

The student's question is about using linear programming to maximize profit in the manufacturing of two types of gadgets under specific constraints. By defining variables, setting up inequalities, and using graphical methods, one can find the optimal production mix of regular and premium gadgets to achieve the highest profit.

Step-by-step explanation:

This question is a linear programming problem from the field of mathematics, where the objective is to maximize profits for a factory that manufactures regular and premium gadgets under specific operational constraints. We need to formulate the problem with algebraic inequalities and then use either graphical methods or simplex methods to identify the optimal number of regular and premium gadgets to produce for maximum profit.

Let the number of regular gadgets be x and the number of premium gadgets be y. We have the following constraints:

• 1x + 2y ≤ 12 (finishing operation hours)
• 2x + 1y ≤ 12 (assembly operation hours)
• x + y ≤ 7 (maximum gadgets per day)
• x, y ≥ 0 (non-negativity constraint)

The profit function to maximize is P = 20x + 30y.

To solve the problem, you graph the inequalities to form a feasible region and find the vertex points. You then calculate the profit function at each vertex point to determine the one with the highest profit. This gives us the optimal number of regular and premium gadgets to manufacture.