Question

COP Appliances has assembly plants in Atlanta and Fort Worth where it produces a variety of kitchen appliances, including a 12-cup coffeemaker and a cappuccino machine. In each hour at the Atlanta plant, 160 of the coffeemakers and 200 of

the cappuccino machines can be assembled, and the hourly cost is $600. In each hour at the Fort Worth plant, 800 of the coffeemakers and 200 of the cappuccino machines can be assembled, and the hourly cost is $1000, CDF Appliances
expects onders each week for at least 36.000 of the coffeemakers and at least 22,000 of the cappuccino machines. How many hours per week should each plant be operated to provide inventory for the orders at minimum cost?
Find the minimum cost.

Answer 1

To find the minimum cost and the number of hours each plant should be operated to meet the demand at minimum cost, you can set up a linear programming problem. Let

�

x represent the number of hours the Atlanta plant is operated per week and

�

y represent the number of hours the Fort Worth plant is operated per week.

The objective is to minimize the total cost, which can be expressed as:

�

=

600

�

+

1000

�

C=600x+1000y

This represents the hourly cost at the Atlanta and Fort Worth plants.

Now, you have two constraints:

The total production of coffeemakers should be at least 36,000:

160

�

+

800

�

≥

36

,

000

160x+800y≥36,000

The total production of cappuccino machines should be at least 22,000:

200

�

+

200

�

≥

22

,

000

200x+200y≥22,000

You also need to consider that the number of hours of operation cannot be negative, so:

�

≥

0

x≥0

�

≥

0

y≥0

Now, you have a linear programming problem with constraints. You can use a solver or graphically solve it to find the values of

�

x and

�

y that minimize the cost

�

C. The solution will give you the minimum cost and the number of hours each plant should be operated.

The minimum cost and the number of hours for each plant can be found using linear programming software or by solving the system of constraints graphically. The solution will provide the optimal values for

�

x and

�

y to minimize the cost while meeting the production demands.