Question 1038585: Two electrician A and B charge 400 and 500 per day respectively. A can service 6 ACs and 4 cooler per day while B can service 10 ACs and 4 cooler per day. For how many days must each be employed so as to service at least 60 ACs and at least 32 Cooler at minimum labour cost? Also calculate the least cost.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! electrician A charges 400 per day.
electrician B charges 500 per day.
electrician A can service 6 AC per day and 4 coolers per day.
electrician B can service 10 AC per day and 4 coolers per day.
assuming each electrician can only do the stated service statistics each day, then the solution would be as follows:
your key measurement is number of days.
your objective function that you want to minimize is dollars per day.
if you let x = the number of days that electrician A works and y equal the number of days that electrician B works, then your objective function becomes .....
400x + 500y
your first constraint is the number of AC that can be serviced.
your first constraint equation is....
6x + 10y >= 60.
this means that the number of days that electrician A needs to service 6 AC per day and the number of days that electrician B needs to service 10 AC per day must be more than or equal to 60 AC total.
your second constraint equation is....
4x + 4y >= 32
this means that the number of days that electrician A needs to service 4 coolers per day and the number of days that electrician B needs to service 4 coolers per day must be more than or equal to 32 coolers total.
a couple of other constraint are assumed.
they are that x and y must be greater than or equal to 0.
you would graph the constraint equations and find the corner points of the feasible region and then analyze the objective function at each corner point to come up with the corner points that generates the least cost.
your graph will contain the following equations.
6x + 10y = 60
4x + 4y = 32
x,y = 0
it will look like this.
your feasible region is the area of the map that is above or on the line of 6x + 10y = 60 and the area of the map that is above or on the line of 4x + 4y = 32 and the area on the map that is on or above the line of x = 0 and the area on the map that is on or to the right of y = 0.
all of these constraints must be satisfied at the same time.
the corner points of that feasible region are shown on the map.
they are (0,8), (5,3), (10,0).
the feasible region is on or above and to the right of the line from (0,8) to (5,3), and on or above and to the right of the line from (5,3) to (10,0) and on or to the right of the line y = 0 and on or above the line x = 0.
it has to meet all those criteria at the same time to be considered in the feasible region.
your minimum or maximum objective function value will be on the corner points of the feasible region.
you need to evaluate the cost function at those points.
you also need to evaluate the constraints at those points in order to ensure all the constraints have been satisfied.
the cost objective function is 400x + 500y = total cost.
at (0,8), the total cost is 0*400 + 8*500 = 4000.
at (5,3), the total cost is 5*400 + 3*500 = 3500.
at (10,0), the total cost is 10*400 + 0*500 = 4000.
your minimum cost is at the point (5,3).
the AC constraint function is 6x + 10y >= 60
at (0,8), the number of AC able to be serviced is 0*6 + 8*10 = 80.
at (5,3), the number of AC able to be serviced is 5*6 + 3*10 = 60.
at (10,0), the number of AC able to be serviced is 10*6 + 0*10 = 60.
the AC constraints are satisfied at all points.
the cooler constraint function is 4x + 4y >= 40
at (0,8), the number of coolers able to be serviced is 0*4 + 8*4 = 32.
at (5,3), the number of coolers able to be serviced is 5*4 + 3*4 = 32.
at (10,0), the number of coolers able to be serviced is 10*4 + 0*4 = 40.
the cooler constraints are satisfied at all points.
your minimum cost point that satisfied all the constraints is at (5,3).
the cost is 3500.
the number of AC that can be serviced is 60.
the number of coolers that can be serviced is 32.
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