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?

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:

Detroit plant can produce:

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?

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

26?

step-by-step explanation:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)