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
+128460 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
