Question

The Cheery Cherry Chair Company makes three types of chairs, the Montegue, the Capulet, and the Verona. Each chair requires assembly, finishing, and a certain amount of wood to make. The assembly area can be used for at most 7 hours per day. To keep him busy, the finishing person needs to work at least 6 hours per day. Producing a Monteguerequires 1 hour of assembly, 2 hour of finishing, and 6 board feet of wood. Producing a Capulet requires 1 hour of assembly, 1 hour of finishing, and 6 board feet of wood. Finally, producing a Verona requires 2 hour of assembly, 1 hour of finishing, and 2 board feet of wood. If the company wants to minimize the amount of cherry wood used, how many should they make of each model? (It is possible to complete a fractional part a chair in a given day; in the long run you end up with completed chairs.)

Number of Montegues =
Number of Capulets =
Number of Veronas =

How many board feet of cherry wood would be used each day?

Answer 1

x₁ = 1.6667 x₂ = 0 x₃ = 2.6667

z(minimum) = 15.3333

Explanation:

Assembly Finishing wood

Montegue (x₁) 1 2 6

Capulet (x₂) 1 1 6

Verona (x₃) 2 1 2

Availability

Assembly area at most 7 hours

Finishing area at least 6 hours

Then

Objective Function

z = 6*x₁ + 6*x₂ + 2*x₃ to minimize

Subject to:

Constraint 1

Assembly area at most 7 hours

x₁ + x₂ + 2*x₃ ≤ 7

Constraint 2

Finishing area at least 6 hours

2*x₁ + x₂ + x₃ ≥ 6

General constraints x₁ ≥ 0 x₂ ≥ 0 x₃ ≥0

Solution after 3 iterations with the help of on-line solver

x₁ = 1.6667 x₂ = 0 x₃ = 2.6667

z(minimum) = 15.3333