Question 209246: 1)K.L. Manufacturing wants to maximize its profit on products A and B. The profit on one unit of Product A is $40, while the profit on Product B is $20. Each unit of Product A requires 10 hours of assembly time and 2 hours of finishing time, while each unit of Product B requires 2 hours of assembly time and 4 hours of finishing time. The departmental capacity (in total hours) is 20,000 for assembly and 31,000 for finishing. What is the maximum profit, and how many of each product should be produced to achieve that profit? Write the objective function and system of linear inequalities and solve it graphically using the method of corners.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! -----
let x = number of units of A
let y = number of units of B
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the constraints are:
profit on unit A is 40 per unit
profit on unit B is 20 per unit
let P = total profit.
this makes P = 40a + 20b
that's your profit equation and that's what you want to maximize.
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product A requires 10 hours of assembly and 2 hours of finishing.
product B requires 2 hours of assembly plus 4 hours of finishing.
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total assembly time available is 20,000 hours.
equation for total hours of assembly time would be:
10x + 2y <= 20,000
that's your first constraint.
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total finishing time available is 31,000 hours.
equation for total hours of finishing time would be:
2x + 4y <= 31000
that's your second constraint.
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you have only 2 constraints as outlined above.
to graph these constraints, you need to put the equations in a form that you can graph them. you do that by solving for y in both equations.
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first equation becomes:
y <= 10,000 - 5x
second equation becomes:
y <= 7750 - .5x
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you graph:
y = 10,000 - 5x
and you graph:
y = 7750 - .5x
and you look for a common area that is underneath both curves.
that area is where all the common solutions to both equations lie.
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if you want to look for a maximum / minimum point, it would be at the corners of the intersections of those lines with each other and with the x and y axis.
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your graph will look like the following:
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the points of intersection would be:
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(0,0)
(2000,0)
(0,7750)
the point of intersection between the two equations.
that point was calculated to be:
(500,7500)
by solving both constraint equations simultaneously (the equality portion, not the inequality portion).
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based on the theory, your maximum profit will be at one of those 4 points.
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your profit equation is 40x + 20y = P
at point (0,0) your profit is 0*40 + 0*20 = $0.00
at point (2000,0) your profit is 2000*40 + 0*20 = $80,000.00
at point (0,7750) your profit is 0*40 + 7750*20 = $155,000
at point (500,7500) your profit is 500*40 + 7500*20 = $170,000
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looks like your maximum profit will be $170,000 when you build 500 units of A and 7500 units of B.
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even though unit A gave you the most profit per unit, it ate up lots of assembly time which didn't allow you to build more than 2000 units. once you built 2000 units of A, you had no more assembly time available for unit B.
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7750 was the maximum units of B you could build because that reached the limit of the hours available for finishing. once you built 7750 units of B, you had no finishing time left for unit A.
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if you built 500 units of A and your built 7500 units of B, then you used up all the assembly time and all the finishing time with no hours left over for either one. since the profit in A was 40 per unit, you were able to get the maximum profit by using this combination.
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the maximum point is at the corners.
i verified that by dropping units of A and / or adding units of B with a corresponding increase / decrease.
you may do the same if you're interested.
keep in mind the constraints, however.
if you add 1 unit of A then you are adding 10 hours of assembly time so you would have to drop 5 units of B to make up the difference (10 = 5*2).
similarly, if you drop 2 unit of A, you can add 1 unit of B because A takes 2 hours of finishing time while B takes 4 hours of finishing time so you need to keep the finishing time required into account.
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