SOLUTION: A steel company has two mills. Mill 1 costs $70,000 per day to operate, and it can produce 400 tons of high-grade steel, 500 tons of medium-grade steel, and 450 tons of low-grade s

Question 1123219: A steel company has two mills. Mill 1 costs $70,000 per day to operate, and it can produce 400 tons of high-grade steel, 500 tons of medium-grade steel, and 450 tons of low-grade steel each day. Mill 2 costs $60,000 per day to operate, and it can produce 350 tons of high-grade steel, 600 tons of medium-grade steel, and 400 tons of low-grade steel each day. The company has orders totaling 100,000 tons of high-grade steel, 150,000 tons of medium-grade steel, and 124,500 tons of low-grade steel. How many days should the company run each mill to minimize its costs and still fill the orders?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = number of days to operate mill 1.
y = number of days to operate mill 2.

your objective function is 70,000 * x + 60,000 * y.

you will evaluate this objective function at each of the corner points of the feasible region on the graph to find the corner point that has the least cost.

your constraint functions are:

400x + 350y >= 100,000 (hi grade steel requirement)
500x + 600y >= 150,000 (medium grade steel requirement)
450x + 400y >= 124,500 (lo grade steel requirement)
x,y >= 0 (number of days can't be negative)

using the desmos.com calculator, you will graph the opposite of these inequalities.

the area on the graph that is not shaded is your region of feasibility.

the corner points of this region are where the minimum cost will be found.

the graph looks like this:

$$$

the corner points are (0,311.25), (210,75), (300,0)

the cost at (0,311.25) is 0 * 70,000 + 311.25 * 60,000 = 18,675,000.
the cost at (310,75) is 210 * 70,000 + 75 * 60,000 = 19,200,000.
the cost at (300,0) is 300 * 70,000 + 0 * 60,000 = 21,000,000.

the minimum cost is when you run mill2 for 311.25 daya.

all the constraint functions need to be met when x = 0 and y = 311.25

400x + 350y = 108937.5 which is > 100,000.
500x + 600y = 186,750 which is > 150,000.
450x + 400y = 124,500 which is e3qual to 124,500.

both x and y are > 0.

all the constraint are met and the cost is minimum when x = 0 and y = 311.25.

that means mill 2 gets a lot of work and mill 1 doesn't get any.

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