Can someone explain how to get the equations from this problem or check if my equations are right:

We have two types of workers tenured and probationary. Our Union contract states that of our 34000 work hours a week at least 20% must be from probationary employees and 40% must be from tenured employees.

*ge is greater than or equal to and le is less than or equal to*

For this part my equation came out to be: 6000x+13600y>(ge)34000

i got this by multiplying 20% to 34000 and 40% to 34000

We know that a tenured employee can produce 1/5 of a shoe per hour and a probationary employee can produce 1/8 of a shoe per hour. Once we know how many shoes we want to produce each week, we need to make sure that the employees we have working can meet that demand.

**there was a part before this where i had to solve for shoes and what i got was 4800 shoes in total 2400 for pair A and 2400 for pair B**

Here all I did was Multiply 1/5*4800=960 and 1/8*4800=600

after this I used this numbers to come up with the equation of: 960x+600y>(ge)4800

Tenured workers make $35 per hour and probationary workers make $20 per hour.

I assume this is the equation needed to solve the problem so It came out to be:

C=35x+20y

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After doing this I have 1 more constrain that I can't seem to figure out and I might have a forth but I don't really know

Guest Jan 22, 2018

#1**+1 **

I swapped x and y from what you had, but the idea is the same

Let x be the hours worked by the probationary employees and y be the hours worked by the tenured employees

So x + y = 34000

At least 20% of the hours must come from the probationay employees and at least 40% come from the tenured employees

So x ≥ 6800 and y ≥ 13600

We must be requiring that the tenured employees produce at least 13600 (1/5) shoes = 2920

So y/5 ≥ 2720

And we must be requiring that the probationary employees must produce at least 6800 (1/8) = 850 shoes

So x/8 ≥ 850

And we want to minimize the cost

So... we want to minimize this objective function : 20x + 35y

Look at the graph here:

https://www.desmos.com/calculator/upq91989qj

We have two possible values that minimize the cost (6800, 27200) and (20,400, 13,600)

Plugging both of these into the objective functions shows that (20,400, 13,600) minimize the cost

CPhill Jan 22, 2018