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A company is making two products A and B. The cost of producing one unit of product A and B is $ 60 and $ 80 respectively. As per the agreement, the company has to supply at least 200 units of product B to its regular customers. One unit of product A requires one machine hours whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirement.

A. Formulate the problem as a linear programming problem.

User Plusplus
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Final answer:

The linear programming problem is to minimize the total cost of production, C = 60x + 80y, with x representing the number of units of product A and y the number of units of product B, subject to the constraints of y ≥ 200, x ≤ 400, and x + y ≤ 500, with both x and y being greater than or equal to zero.

Step-by-step explanation:

To formulate this problem as a linear programming problem, we need to define variables, the objective function, and the constraints based on the provided information. Let's define x as the number of units of product A and y as the number of units of product B. The objective function would be to minimize the total cost of production, which can be represented as C = 60x + 80y, where 60 and 80 represent the costs of producing one unit of product A and B respectively.

There are several constraints to consider:

The company has to supply at least 200 units of product B, so we have y ≥ 200.

The total machine hours for product A are limited to 400 hours, which gives us the constraint x ≤ 400 since one unit of product A requires one machine hour.

One unit of each product requires one labor hour each, with a total of 500 labor hours available, leading to the constraint x + y ≤ 500.

Also, we must consider that the number of products manufactured cannot be negative, thus x ≥ 0 and y ≥ 0.

The linear programming problem can therefore be summarized as follows:

Objective Function:

Minimize C = 60x + 80y

Subject to Constraints:

y ≥ 200

x ≤ 400

x + y ≤ 500

x ≥ 0

y ≥ 0

User KobeJohn
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