Question

A tire company has plants in Chicago and Detroit. The Chicago plant can make 600 radial and 100 standard tires per day. The Detroit plant can make 300 radial and 100 standard tires per day. It costs $4,000 per day to operate the Chicago plant and $3,000 per day to operate the Detroit plant. The company has a contract to make at least 24,000 radial and 5000 standard tires.

How many days should each plant be scheduled to minimize operating costs?

Answer 1

To minimize operating costs, the Chicago plant should work 30 days and the Detroit plant should work 20 days

Explanation:

1. Let's review the information given to us to answer the question correctly:

Chicago plant can produce:

• 600 radial and 100 standard tires per day
• Cost of daily operation = $ 4,000

Detroit plant can produce:

• 300 radial and 100 standard tires per day
• Cost of daily operation = $ 3,000

Contract to make at least 24,000 radial and 5000 standard tires.

2. How many days should each plant be scheduled to minimize operating costs?

• Let C to represent the number of days the Chicago plant to work
• Let D to represent the number of days the Detroit plant to work

We can write our system of inequalities, this way:

100C + 100D ≥ 5,000

600C + 300D ≥ 24,000

Isolating C:

100C ≥ 5,000 - 100D

C ≥ 50 - D (Dividing by 10 at both sides)

Substituting C and solving for D in the 2nd inequality:

600(50 - D) + 300D ≥ 24,000

30,000 - 600D + 300D ≥ 24,000

-300D ≥ 24,000 - 30,000

-300D ≥ - 6,000

D ≥ -6,000/-300

D ≥ 20

Solving for C in the 1st inequality:

100C + 100 * 20 ≥ 5,000

100C + 2,000 ≥ 5,000

100C ≥ 5,000 - 2,000

100C ≥ 3,000

C ≥ 3,000/100

C ≥ 30

In consequence, the total costs of operating the plants would be:

30 * 4,000 + 20 * 3,000

120,000 + 60,000 = $ 180,000

Answer 2

To minimize the cost, the Chicago plant should run 30 days and the Detroit plant should run 20 days.

Explanation:

This is a linear programming question, in which there is a function to optimize (minimize cost) and restrictions, all linear functions.

The objective function is Cost:

`C=4000H+3000D`

C: cost function

H: days of operation for Chicago plant

D: days of operation for Detroit plant

The restrictions are:

- Minimum requirements for radial tires:

`600*H+300*D\geq24,000`

- Minimum requirements for standard tires:

`100*H+100*D\geq5,000`

- Days are positive integers

`H,D\geq0`

We have 3 points in which 2 of the constraints are saturated. In one of this three points is the minimum cost.

We will evaluate them to find the minimum cost:

Point 1: H=0, D=80

`C=4,000*H+3,000*D\\\\C=4,000*0+3,000*80=240,000`

Point 2: H=50, D=0

`C=4,000*H+3,000*D\\\\C=4,000*50+3,000*0=200,000`

Point 3: H=30, D=20

`C=4,000*H+3,000*D\\\\C=4,000*30+3,000*20=120,000+60,000=180,000`

The third point minimizes the cost.

To minimize the cost, the Chicago plant should run 30 days and the Detroit plant should run 20 days.