Respuesta :
Answer:
To minimize the cost, the Chicago plant should run 30 days and the Detroit plant should run 20 days.
Step-by-step 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:
[tex]C=4000H+3000D[/tex]
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:
[tex]600*H+300*D\geq24,000[/tex]
- Minimum requirements for standard tires:
[tex]100*H+100*D\geq5,000[/tex]
- Days are positive integers
[tex]H,D\geq0[/tex]
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
[tex]C=4,000*H+3,000*D\\\\C=4,000*0+3,000*80=240,000[/tex]
Point 2: H=50, D=0
[tex]C=4,000*H+3,000*D\\\\C=4,000*50+3,000*0=200,000[/tex]
Point 3: H=30, D=20
[tex]C=4,000*H+3,000*D\\\\C=4,000*30+3,000*20=120,000+60,000=180,000[/tex]
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.

Answer:
To minimize operating costs, the Chicago plant should work 30 days and the Detroit plant should work 20 days
Step-by-step 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