Question 171053: 1) The Rose Model Company is planning to market a variety of electric racing car sets. Each set will contain at least 8 sections of curved tracks and 4 sections of straight tracks. No set will contain more than 36 sections in all or more than 20 sections of either type. If the company makes a profit of $0.40 on each straight track and $0.65 on each curved track, what combination of track sections will be most profitable for the company?
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! let x = number of straight tracks.
let y = number of curved tracks.
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equation you are trying to maximize is:
p = .4x + .65y
where p is profit.
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your constraints are as follows:
each set will contain at least 8 sections of curved tracks and 4 sections of straight tracks.
x >= 4
y >= 8
no set will contain more than 36 sections in all.
x + y <= 36
no set will contain more than 20 sections of either type.
x <= 20
y <= 20
since x >= 4 and <= 20, this can be rewritten as:
4 <= x <= 20
since y >= 8 and <= 20, this can be rewritten as:
8 <= y <= 20
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take the equation of x + y <= 36 and solve for y as an equality.
you get x + y = 36
subtract x from both sides:
y = -x + 36
this equation can be graphed in this form easily.
you would want to graph the following equations:
y = -x + 36
x = 4
x = 20
y = 8
y = 20
the intersections of these equations will help you to find the intersections where the minimum or maximum values of your objective equation are located.
the profit equation is your objective equation.
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a picture of this graph can be found at http://www.geocities.com/gonzo89p.
look for 171053 picture.
it should be up there no later than 1 hour after you receive this answer. hopefully it will be up there sooner.
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looking at the graph you will see that you have 5 intersection points of interest.
the intersections are solved for as follows:
1. intersection of x = 4 and y = 20 is: (4,20)
2. intersection of x = 4 and y = 8 is: (4,8)
3. intersection of x = 20 and y = 8 is: (20,8)
4. intersection of y = -x + 36, and x = 20 is: (20,16)
5. intersection of y = -x + 36, and y = 20 is: (16,20)
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the area of interest on that graph is:
x >= 4
x <= 20
y >= 8
y <= 20
y <= -x + 36
that's the shaded area in the picture.
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the maximum / minimum points of the equation of interest will be at the intersections shown.
you need to solve each one using the values of x and y from those intersections and then find which is the maximum one.
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you equation of interest is the profit equation:
p = .4x + .65y
use this to solve for each set as follows:
(4,20) = 14.6
(4,8) = 6.8
(20,8) = 13.2
(20,16) = 18.4
(16,20) = 19.4
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it appears that the maximum profit will be at the intersection point of (16,20).
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that's when x = 16 and y = 20.
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logically this makes sense.
y give the most profit.
y can't be greater than 20.
so sell 20 * y and the rest x.
since both can't be greater than 36, that makes the maximum x = 16.
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careful to make sure that x = 16, and y = 20 is a valid point that satisfies the conditions.
since all your constraint equations were <= or >=, then the actual point itself is valid.
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i will post the graph on the website right away.
be patient if it's not there when you look.
wait a half hour and it will be the next time you look unless i ran into problems.
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