Question 344076
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Let *[tex \Large x] represent the number of miles traveled.


Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C_1(x)\ =\ 0.20x\ +\ 20]


Represents the total cost in dollars of renting from Company #1 for *[tex \Large x] miles.


Whereas:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C_2(x)\ =\ 0.10x\ +\ 38]


Represents the total cost in dollars of renting from Company #2 for *[tex \Large x] miles.


Clearly, for small values of *[tex \Large x], *[tex \Large \ \ C_1] is less expensive.  Substitute 1 mile for *[tex \Large x] in each of the functions and see which is smaller, i.e. less expensive.


Also, for large values of *[tex \Large x], *[tex \Large \ \ C_2] becomes less expensive at some point.  Substitute 1000 for *[tex \Large x] in each of the functions and see which is smaller, i.e. less expensive.


Somewhere between 1 mile and 1000 miles is a point where the cost of both becomes equal.  This is called the break-even point.  Mileage below the breakeven point makes *[tex \Large C_1] less, above the breakeven point makes *[tex \Large C_2] less.


The breakeven point is where the two costs are equal:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C_1(x)\ =\ C_2(x)]


But we can replace each side of this equation with the expressions they are equal to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.20x\ +\ 20\ =\ 0.10x\ +\ 38]


Solve for *[tex \Large x].


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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