Question 35573This question is from textbook 0-618-49281-x
: Thanks for your help---
A company makes two models of a product. The times (in hours) required for assembling, painting, and packaging are as follows:
assembly (model a takes 2.5) (model b takes 3)
painting (moel a takes 2)(model b takes 1)
packaging (model a takes .75)(model b takes 1.25)
The total times available for assembling, paiting, and packaging are 4000 hours, 2500 hours, and 1500 hours, respectively. The profits per unit are $50 for model A and $52 for model B. How many of each should be made to obtain a maximum profit? I have a final Wed if you can help thank you thank you :)
This question is from textbook 0-618-49281-x
Found 2 solutions by narayaba, stanbon:
Answer by narayaba(40)   (Show Source):
You can put this solution on YOUR website! This is an optimization problem also known as linear programming
Let x be the number of models of type a
Let y be the number of models of type b
time needed for assembling x number of model a is 2.5x
time needed for assembling y number of model b is 3y
Total time available for assembling is 4000
the first inequality is the total assembly time for both model a and model b shoulde be less than or equal to 4000 which can expressed as
2.5x + 3y <=4000 which is same as 5x + 6y <=8000 (1)
similarly
time needed for painting x number of model a is 2x
time needed for painting y number of model b is y
Total time available for painting is 2500
the second inequality for the painting time can be expressed as
2x + y <=2500 (2)
time needed for packaging x number of model a is 0.75x
time needed for packaging y number of model b is 1.25y
Total time available for painting is 2500
and finally the third inequality from the packaging time can be written as
0.75x + 1.25y <=1500 which is same as 3x + 5y <=6000 (3)
The price for model a is $50 and price model for b is $52
The total price is 50x + 52y which needs to be maximized with the above formulated constraints.
maximize 50x + 52y subject to
5x + 6y <=8000
2x + y <=2500
3x + 5y <=6000
and also that x>=0 and y>=0 since it is not possible to produce negative number of model a and model b
to find the solution the first step is to graph all the inequalities or constraints
Th region for inequality 1 is the region that contains the point (0,0) since (0,0) satisfies the inequality (1) 0 < 8000 So the region beneath and to the left of the line colored maroon is the region corresponding to the inequality 1
similarly the region for the inequality (2) is the one to the left and beneath of line colored green finally the region for the inequality (3) is the one to the left and beneath of line colored blue
and the solution is the region formed by the intersection of all the ineqaulities along with x>=0 and y>=0
The cordinates of intersecting region is plugged into the equation 50x + 52y and the coordinate that gives the maximum vlue is chosen as the optimum values of x and y. For this problem the cordinates of the intersecting region are (0,0)(0,1200) (4000/7, 6000/7) (1000,500) (1250,0). Plugging in each of the coordinates into 50x + 52y
we get
(0,0)= 0
(0,1200) = 52*1200 = 62400
(4000/7, 6000/7) = 50*(4000/7) + 52*(6000/7) = 73142.8
(1000,500) = 50*1000 + 52*500 = 76000
(1250,0) = 50*1250 + 50*0 = 62500
The coordinate that gives the maximum value for 50x + 52y is (1000,500) and the value is $76000. Hence the company must produce 1000 quantities of model a and 500 of model b to get a maximum profit of $76000
It requires little more usage of graphing to completley explain this problem. I am not sure of how to shade the regions with available tools to make it more simple for you to understand and follow the material presented in here. It would have been nice to have a white board to explain the details. Hope you get the idea.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Are you sure you have the data correct?
Inequalities based on your posted data give:
2.5a+3b<=4000
2a+b<=2500
0.5a+1.25b<=1500
Plotting these you get a triangle
with two vertices belog the "a" axis indicating
negative production of model "b".
Only one axis has a meaningful interpretation,
i.e. (1000,0)
Indicating production of 1000 model "a" and no "b"
That would give a profit of 1000(50) = $50,000
Does this sound reasonable?
Cheers,
stan H.
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