Question

A furniture manufacturer makes two types of furniture: chairs and sofas. The production of the sofas and chairs requires three operations: carpentry, finishing, and upholstery. Manufacturing a chair requires 3 hours of carpentry, 9 hours of finishing, and 2 hours of upholstery. Manufacturing a sofa requires 2 hours of carpentry, 4 hours of finishing, and 10 hours of upholstery. The factory has allocated at most 66 labor hours for carpentry, 180 labor hours for finishing, and 200 labor hours for upholstery. The prfit per chair is $90 and the profit per sofa is $75. The manufacturer wants to know how many chairs and how many sofas should be produced each day to maximize the profit.
Formulate a linear programming (LP)problem you would use to find a solution.

Answer 1

Answer:

Chair  x₁ = 10

Sofa   x₂ = 18

z(max)  = 2250 $

Step-by-step explanation:

Information:

Operation-Profit             Carpentry         Finishing      Upholstery     Profit

                                           hours                 hours              hours             $

Chair (x₁)                                3                        9                      2                90

Sofa  (x₂)                                 2                        4                     10               75

Objective Function z:

z =  90*x₁ + 75*x₂

Constrains:

1.-66 hours of carpentry available

3*x₁ + 2*x₂ ≤ 66

2.- 180  hours of finishing available

9*x₁  +  4*x₂ ≤  180

3.- 200 hours of upholstery

2*x₁  +  10*x₂  ≤ 200

Model:

z = 90*x₁ + 75*x₂    to maximize

Subject to:

3*x₁ + 2*x₂ ≤ 66

9*x₁  +  4*x₂ ≤  180

2*x₁  + 10*x₂  ≤ 200

x₁ >= 0  x₂ >= 0

From simplex method solver we get the answer

x₁ = 10         x₂  = 18       z = 2250

Answer 2

The linear programming model is:

• Maximize $\mathbf{Z = 90x + 75y}$
• Subject to $\mathbf{3x + 2y \le 66}$, $\mathbf{9x + 4y \le 180}$, $\mathbf{2x + 10y \le 200}$, $\mathbf{x,y \ge 0}$

Represent sofa with y and chairs with x.

The given parameters are:

Carpentry:

$\mathbf{Chair =3\ hours}$

$\mathbf{Sofa =2\ hours}$

$\mathbf{Available = 66\ hours}$

So, we have:

$\mathbf{3x + 2y \le 66}$

Finishing

$\mathbf{Chair =9\ hours}$

$\mathbf{Sofa =4\ hours}$

$\mathbf{Available = 180\ hours}$

So, we have:

$\mathbf{9x + 4y \le 180}$

Upholstery

$\mathbf{Chair =2\ hours}$

$\mathbf{Sofa =10\ hours}$

$\mathbf{Available = 200\ hours}$

So, we have:

$\mathbf{2x + 10y \le 200}$

The profit is given as: $90 per chair, and $75 per sofa.

So, the objective function is:

$\mathbf{Z = 90x + 75y}$

Hence, the linear programming model is:

Maximize $\mathbf{Z = 90x + 75y}$

Subject to

$\mathbf{3x + 2y \le 66}$

$\mathbf{9x + 4y \le 180}$

$\mathbf{2x + 10y \le 200}$

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

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