Question 1191606: A manufacturer produces a business calculator and a graphing calculator. each calculator is assembled in two sets of operations, where each operation is in production 8 hours during each day. the average time required for a business calculatorin the first operation is 3 minutes, and 6 minutes is required in the second operation. the graphing calculator averages 6 minutes in the first operation and 4 minutes in the second operation. all calculators can be sold the profit for business calculator is 8$ and the profit for a graphing calculatoris $10. how many of each type of calculator should be made each day in order to maximize profit.
I know that one equation will include X= the business calculator and y= the graphing calculator.
x+y<=8 hours?
9x+10y<=?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i think you need to convert hours to minutes, since the operations are in minutes.
alternatively, you can convert minutes to hours, but it seems to make more sense to convert hours to minutes.
8 hours * 60 minutes per hour = 480 minutes.
let x = the number of business calculators.
let y = the number of graphing calculators.
since the profit per business calculator is 8 dollars and the profit per graphing calculator is 10, then your objective function is profit = 8x + 10y.
this is the equation that will be used to evaluate the corner points of the feasible region.
you have two operations required to assemble a calculator.
for the business calculator, the first operation requires 3 minutes and the second operation requires 6 minutes.
for the graphing calculator, the first operation requires 6 minutes and the second operation requires 4 minutes.
each of these operations can be used for up to 8 hours each day.
translated to minutes, each of those operations can be used for up to 480 minutes each day.
your constraint inequalities are:
3x + 6y <= 480
6x + 4y <= 480
the first inequality says that 3 minutes times the number of business calculators and 6 minutes times the number of graphing calculators must total to less than or equal to 480 minutes each day for the first operation.
the second inequality says that 6 minutes times the number of business calculators and 4 minutes times the number of graphing calculators must total to less than or equal to 480 minutes each day for the second operation.
two additional constraints are that the number of business calculators and the number of graphing calculator can't be negative.
those inequalities are:
x >= 0
y >= 0
using the desmos.com calculator, you would graph the opposite of these inequalities.
the area on the graph that is not shaded will be your region of feasibility.
the lines of the equations themselves will also be in the region of feasibility if the inequalities are <= or >=, rather than < or >.
to summarize.
your objective function is:
8x + 106 = profit
your constraint inequalities are:
3x + 6y <= 480
6x + 4y <= 480
x >= 0
y >= 0
you are graphing the constraints and you are evaluating each corner point with the objective function.
the graph is shown below.
the corner points on the graph are (0,80), (40,60), (80,0).
these are coordinate points in (x,y) format.
x is the number of business calculators.
y is the number of graphing calculators.
you would evaluate the objective function of profit = 8x + 10y at each of the corner points of the feasible region.
the corner point with the maximum profit will be at (40,60).
all the constraints need to be also satisfied, not just some.
3x + 6y <= 480 becomes 120 + 360 <= 480 which becomes 480 <= 480 which is true.
6x + 4y 480 becomes 240 + 240 <= 480 which becomes 480 <= 480 which is also true.
both x and y are greater than or equal to 0.
since all the constraints are satisfied at the maximum profit point, then the maximum profit will be when 40 business calculator and 60 graphing calculators are assembled and sold.
i'll be available to answer any questions or concerns you have about this analysis.
theo
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