SOLUTION: How do I solve the following problem?
A boat manufacturer makes fishing boats that are sold for a profit of $500. each, and canoes that are sold for a profit of $400 each. Each f
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Question 220119: How do I solve the following problem?
A boat manufacturer makes fishing boats that are sold for a profit of $500. each, and canoes that are sold for a profit of $400 each. Each fishing boat requires 100 assembly hours and 25 finishing hours, while each canoe requires 75 assembly hours and 50 finishing hours. How many fishing boats and how many canoes should be made to maximize the manufacturer's profit? Let x= the number of fishing boats and y=the number of canoes. thanks for your help.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = number of fishing boats
y = number of canoes
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profit per fishing boat is $500
profit per canoe is $400
profit equation would be:
p = 500*x + 400*y
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your constraints appear to be:
total assembly hours = 100*x + 75*y
total finishing hours = 25*x + 50*y
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To keep these straight, it helps to create a table that would look something like:
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craft_________________assembly time required__________finishing time required
fishing boat_________________100 hours______________________25 hours
canoe______________________75 hours______________________50 hours
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you appear to be missing some constraints like total assembly hours <= ... and total finishing hours <= ...
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I will insert some hours and solve for that.
You can then replace whatever I have with whatever you have and solve the problem for yourself.
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Without those constraints, the problem is already solved because if he has unlimited time to assemble and finish then he should sell all fishing boats because he gets more profit per fishing boat.
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I'll put in some constraints and we'll see how it works out.
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Total hours available for assembly are <= 1000
Total hours available for finishing are <= 500
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Our constraints become:
100x + 75y <= 1000
25x + 50y <= 500
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Summary:
Profit equation is 500*x + 400*y
Constraint equation number 1 is 100x + 75y <= 1000
Constraint equation number 2 is 25x + 50y <= 500
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We graph the constraint equations.
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We do that by solving for y in both equations.
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We solve for equality, not for inequality.
This allows us to graph the equations.
Inequality is handled by looking at the area above or below the graph of the equations.
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Constraint equation 1 becomes y = (1000-100x)/75
Constraint equation 2 becomes y = (500-25x)/50
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graph of these equations is shown below:
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Picture of this graph with the area of possible solutions can be viewed by clicking on the following hyperlink.
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Graph of Equations and Area of Possible Solutions
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The area of possible solution is bounded by:
The x-axis because the value of y, which is number of canoes, has to be >= 0.
The y-axis because the value of x, which is number of fishing boats, has to be >= 0.
The line y = (1000-100x)/75 because the number of hours to assemble a fishing boat (x) and a canoe (y) has to be <= 1000.
the line y = (500-25x)/50 because the number of hours to finish a fishing boat (x) and a canoe (y) has to be <= 500.
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Remember that x equals the number of fishing boats, and y equals the number of canoes.
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The equation for total assembly time is 100*x + 75*y <= 1000
We converted this to:
y <= (1000-100*x)/75 in order to graph it, and then we graphed:
y = (1000-100*x)/75
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When y equals the value of this equation, y will be ON the line.
When y is smaller than the value of this equation, y will be BELOW the line.
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Our area of possible solutions for this equation are thye points ON the line and the points BELOW the line.
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Same for the other equation.
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The possible solutions for the x-axis are the value of y ON the x-axis and the values of y ABOVE the x-axis.
The possible solutions for the y-axis are the values of x ON the y-axis and the values of x TO THE RIGHT OF the y-axis.
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The rules of linear programming states that the maximum/minimum values will be at the intersection of the boundaries of the constraints.
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This means we have to look at the points:
(0,10)
(0,0)
(10,0)
(4,8)
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We apply these points to our revenue equation of 500*x + 400*y
Point (0,10) yields 500*0 + 400*10 = $4,000.00
Point (0,0) yields 500*0 + 400*0 = $0.00
Point (10,0) yields 500*10 + 400*0 = $5,000.00
Point (4,8) yields 500*4 + 400*8 = $2,000.00 + $3,200 = $5,200
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Our maximum profit can be attained when we build and sell 4 fishing boats and 8 canoes.
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Why couldn't we get more profit when we build all fishing boats and no canoes?
It would seem that a greater profit on each fishing boat should result in building all fishing boats.
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Here's what happened.
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When we built 10 fishing boats, we used up all the assembly time available so we had no assembly time left over for the canoes.
That means we had to stop, even though we had some finishing time left for the canoes.
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When we built 4 fishing boats, we still have assembly time left over for 8 canoes. 4*100 = 400 subtracted from 1000 = 600 divided by 75 hours for each canoe equals 8 canoes able to be assembled with the time left over. This resulted in a total of 12 craft being assembled and finished and sold (presumption is you are able to sell all that you build). Even though the profit for each canoe was less than the profit for each fishing boat, more craft total yielded more total profit. 4*500 + 8*400 = 5200 while 10*500 = 5000.
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