Question

A factory manufactures chairs and tables, each requiring the use of three operations: cutting, assembly, and finishing. The first operation can use at most 40 hours; the second at most 42 hours; and the third at most 25 hours. A chair requires 1 hour of cutting, 2 hours of assembly, and 1 hour of finishing; a table needs 2 hours of cutting, 1 hour of assembly, and 1 hour of finishing. If the profit is $20 per unit for a chair and $30 for a table, what is the maximum revenue? Round your answer to the nearest whole number. Do not include a dollar sign or comma in your answer.

Answer 1

Answer:

z(max) = 650 $

x₁ = 10 units

x₂ = 15 units

Step-by-step explanation:

That is a linear programming problem, we will use a simplex method to solve it

Formulation:

Let´s call  x₁  number of chairs   and x₂ number of tables then :

Item              (in hours)     cutting       assembly      finishing        Profit ($)

Chairs (x₁)                              1                   2                     1                      20

Tables (x₂)                              2                   1                      1                      30

Availability                           40                 42                   25

Objective Function

z  =  20*x₁   +  30x₂   ( to maximize) subject to:

x₁  +  2x₂   ≤  40

2x₁  + x₂    ≤  42

x₁ + x₂     ≤    25

x₁  ,   x₂  >= 0

Using excel or any other software we find:

z(max) = 650

x₁ = 10

x₂ = 15

Answer 2

The chairs and tables manufactured by the factory is an illustration of linear programming, where the maximum revenue is 674

Let x represent chairs, and y represent tables

So, the given parameters are:

Cutting:

• Chairs: 1 hour
• Table: 2 hours
• Hour available: 40

So, the constraint is:

$\mathbf{x + 2y \le 40}$

Assembly:

• Chairs: 2 hours
• Table: 1 hour
• Hour available: 42

So, the constraint is:

$\mathbf{2x + y \le 42}$

Finishing:

• Chairs: 1 hour
• Table: 1 hour
• Hour available: 25

So, the constraint is:

$\mathbf{x + y \le 25}$

The unit profit on the items are:

• Chairs: $20
• Table: $30

So, the objective function to maximize is:

$\mathbf{Max\ z = 20x + 30y}$

And the constraints are:

$\mathbf{x + 2y \le 40}$

$\mathbf{2x + y \le 42}$

$\mathbf{x + y \le 25}$

$\mathbf{x,y \ge 0}$

Using graphical method (see attachment for graph), we have the following feasible points:

$\mathbf{(x,y) = \{(10,15),\ (17,8),\ (14.67, 12.67)\}}$

Calculate the objective function using the feasible points.

$\mathbf{z = 20 \times 10 + 30 \times 15}$

$\mathbf{z = 650}$

$\mathbf{z = 20 \times 17 + 30 \times 8}$

$\mathbf{z = 580}$

$\mathbf{z = 20 \times 14.67+ 30 \times 12.67}$

$\mathbf{z = 673.5}$

Approximate

$\mathbf{z = 674}$

Hence, the maximum revenue is 674

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