Question

A factory manufactures chairs, tables and bookcases each requiring the use of three operations: Cutting, Assembly, and Finishing. The first operation can be used at most 600 hours; the second at most 500 hours; and the third at most 300 hours. A chair requires 1 hour of cutting, 1 hour of assembly, and 1 hour of finishing; a table needs 1 hour of cutting, 2 hours of assembly, and 1 hour of finishing; and a bookcase requires 3 hours of cutting, 1 hour of assembly, and 1 hour of finishing. If the profit is $20 per unit for a chair, $30 for a table, and $25 for a bookcase, how many units of each should be manufactured to maximize profit?

Answer 1

Using linear programming, the factory should manufacture 200 chairs, 150 tables, and 50 bookcases to maximize the profit.

Forming the linear programming objective and constraints functions:

Cutting Assembly Finishing Profits

Maximum operating hours 600 500 300

Chair requirements 1 1 1 $20 per unit

Table requirements 1 2 1 $30 per unit

Bookcase requirements 3 1 1 $25 per unit

Let the number of chairs = x

Let the number of tables = y

Let the number of bookcases = z

Objective function: maximize 20x + 30y + 25z

Subject to the constraints:

1x + 1y + 3z ≤ 600 (cutting operation)

1x + 2y + 1z <= 500 (assembly operation)

1x + 1y + 1z ≤ 300 (finishing operation)

Additionally, x, y, and z need to be non-negative (x >= 0, y >= 0, z >= 0).

Using a linear programming solver, we can input the objective function and constraints to find the optimal values for x, y, and z.

The optimal values are:

x = 200 (number of chairs)

y = 150 (number of tables)

z = 50 (number of bookcases)

Thus, the factory should manufacture 200 chairs, 150 tables, and 50 bookcases to maximize the profit.