SOLUTION: I need help. I have been trying to work this problem for several days and I am confused. This is the problem: The Johnson family is planning a weeklong vacation in California and

Question 169697: I need help. I have been trying to work this problem for several days and I am confused. This is the problem: The Johnson family is planning a weeklong vacation in California and needs to rent a car. They don't know exactly how far they will drive, but they estimate between 400 and 800 miles. Here are the options:
Weekly Rates: $329 per week, unlimited mileage or $219 per week, plus 12 cents per mile.
Daily Rates: $50 per day, unlimited mileage or $40 per day, plus 3 cents per mile.
Write the total week's rental car cost as a function of the number of miles driven in a week for each of the options. The costs of each option must also be represented on a data table and then graphed to show the cost of each option.

Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
you have 4 options.
you need to outline how much each option will cost per week.
assume a week is 7 days without any other information to go on.
they will travel between 400 and 800 miles (given).
option 1:
$329 + $0.0 per mile
option 2:
$219 + $.12 per mile
option 3:
7*$50 = $350 + $0.0 per mile
option 4:
7*$40 = $280 + $.03 per mile
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easiest way to do this is to create a function and graph it.
i'll make the functions a,b,c,d
a(x) = 329 + 0*x
b(x) = 219 + .12*x
c(x) = 350 + 0*x
d(x) = 280 + .03*x
graph of all 4 equations follows:
look below the graph for further comments.
negative numbers on the graph are only there so the x and y axis will show up. it won't otherwise.
you can figure out which line is which by looking at the starting values.
on this graph, the x axis is the number of miles traveled and the y axis is the cost.

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assuming you will not travel more than 800 miles, the two plans with 0 cost per mile are too expensive relative to the other plans. they only become competitive over 800 miles traveled.
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the cheapest plan in the beginning is the one that start off at $219. it's the best up to about 600 miles when the plan that starts off at $280 becomes cheaper.
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you could actually solve for the crossover points.
take the plans that start off at $219 and $280.
those are:
b(x) = 219 + .12x
d(x) = 280 + .03x
just make them equal to each other and solve for x.
219 + .12x = 280 + .03x
subtract .03x from both sides:
219 + .12x - .03x = 280
subtract 219 from both sides:
.12x - .03x = 280 - 219
combine like terms:
.09x = 61
divide both sides by .09:
x = 61 / .09 = 677.777777...... miles.
that's the crossover point when both these plans are equal.
it's hard to see on the graph but it's in the general vicinity of 679 miles on the graph.
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if you want to plot data points, you may.
takes miles from 400 to 800 in 50 mile chunks and solve for each equation at each of those points.
for example:
1 point is 400 miles
at 400 miles you'll get these numbers:
a(x) = 329 + 0*400 = 329
b(x) = 219 + .12*400 = 219 + 48 = 267
c(x) = 350 + 0*400 = 350
d(x) = 280 + .03*400 = 280 + 12 = 292
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you would need to do this every 50 or 100 miles, whichever you prefer. since it's less work, do it every 100 miles.
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looking at the graph is easier.
since they're all straight lines, you can actually plot at 400 miles and at 800 miles and draw a straight line in between.
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