Question 1062038: Good evening, can you please help me solve the 3rd one? Thank you very much!
(1) An office manager needs to staff the office. She hires full-time employees at $18 per our and part-time employees at $12 per hour. Write an objective function that represents the total cost (in $) to staff the office with x full-time employees and y part-time employees for 1 hr.
ANSWER: Total cost: z = 18x + 12y
(2) Refer to 1. Suppose that the office manager needs at least 20 employees, but not more than 24 full-time employees. Furthermore, to make the office run smoothly, the manager knows that the number of full-time employees must always be greater than or equal to the number of part-time employees. Write a system of inequalities that represents the constraints on the number of full-time employees x and the number of part-time employees y.
ANSWER: x+y≥20
x≤24
x≥y
(3) Refer to no's 1 and 2. The office manager needs at least 20 employees, but not more than 24 full-time employees. Furthermore, to make the office run smoothly, the manager knows that the number of full-time employees must always be greater than or equal to the number of part-time employees. If she pays full-time employees $18 per hour and part-time employees $12 per hour, determine the number of full-time and part-time employees she should hire to minimize total labor cost per hour.
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52794)   (Show Source):
You can put this solution on YOUR website! .
(2) You should amplify your inequalities in the n(2) with these two inequalities: x >= 0, y >= 0.
(as I said in my preceding post).
(3) Your area to apply the linear programming method is shown in the figure below:
Lines x + y >= 20 (black), x = 24 (red), y = x (blue)
This area is the quadrilateral in the first quadrant, restricted by the straight line x+y >= 20 (black),
straight line y = x (blue), vertical line x = 24 (red) and horizontal line y = 0.
According to the linear programming method, you should calculate your objective function in four vertices of the quadrilateral
and choose the point where the objective function is minimal.
This point is your solution.
Together with the minimal value of the objective function.
That's all.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you graph the constraints and then you evaluate the objective function at the corners of the feasible region.
in the attached graph, the feasible region is the region that is NOT shaded.
here's the graph:
the corner points are:
(10,10)
(20,0)
(24,0)
(24,24)
evaluate the objective function at these corner points and the minimum cost is when 10 full time employers are hired and 10 part time employees are hired.
the objective function is the cost function of z = 18x + 12y
at the coordinate point of (10,10), the cost is 180 + 120 = 400.
the constraints at (10,10) are satisfied.
x + y >= 20
x <= 24
x >= y
the total constraints are:
x + y >= 20
x <= 24
x >= y
x >= 0
y >= 0
the last two are necessary because neither x nor y can be less than 0.
in the graph, i shaded the regions that do NOT satisfy the constraints.
what is left is the region that DOES satisfy the constraints.
you still need to pay attention to the original constraints since that tells you what your answer can be.
for example, if the constraint was x + y > 20, then (10,10) would not satisfy that constraint because it is not greater than 20.
(10,10) would still be a marker that tells you the area where the answer might lie.
(11,10) is still in the region of feasibility, as is (10,11).
either one of those would suffice if the requirement was > rather than >=.
in fact (10,11) would be the least cost solution in that case, since the pay of the temporary employee is less than the pay of the regular employee.
the software to graph that i used is at www.desmos.com
here's a reference on graphing linear inequalities.
http://www.purplemath.com/modules/ineqgrph.htm
note that they tell you to shade the region of feasibility and that to use dashed lines if the inequality does not contain an equal as part of it.
i did the opposite only because i was not manually creating the graph and found that not shading the region of feasibility allowed it to show up better when using desmos software to generate the graph.
the general idea is that you graph the region of feasibility and then look for the corner points of that region.
your minimum / maximum solution, if there is one, should be at those corner points.
here's another reference that addresses corner points.
http://www.purplemath.com/modules/linprog.htm
|
|
|
