1. Sep 9, 2009

### Loppyfoot

1. The problem statement, all variables and given/known data

4. Sep 9, 2009

### Loppyfoot

I got 72,000 miles for where they are equal. So, after 72,001 miles would make Car B the best choice.

Do you know if there is a faster way to complete this problem?

5. Sep 9, 2009

That's the same answer I got. I don't know about a faster way. I initially tried just figuring out how to get the $1500 difference, but then did the sketch/plot, and realized that I was neglecting the initial investment cost. If you think about it some, and keep checking your alternate methods' answers, you may find a good trick. Good luck! 6. Sep 9, 2009 ### Loppyfoot I found that if I plugged in the formula: CostA = ((1.25x)/20) + 4000 CostB = ((1.25x)/30) + 5500 into my calculator, and then made a table, plugging in values of x; in factors of 10000; it sped it up a little bit. 7. Sep 9, 2009 ### Elucidus Conceptually one can approach the problem in much the same way as the OP. Car B costs$1500 more than A but gets better gas milleage. What is important to consider is the relative savings between the cars. The cost per mile for A is (1.25)/20 while B is (1.25)/30. The difference is 1/48. So it boils down to how many many miles at $1/48 per mile savings must one drive in order to save$1500? The answer is 48 x 1500 = 72,000 miles.

Whats really happening here is solving the system

$$\left\{ \begin{array}{c} y=4000 + \frac{1.25}{20}x \\ \\ y=5500 + \frac{1.25}{30}x \end{array}$$

which results in the equation and solution:

$$5500 + \frac{1.25}{30}x = 4000 + \frac{1.25}{20}x$$

$$5500 - 4000 = \frac{1.25}{20}x - \frac{1.25}{30}x$$

$$1500 = \left( \frac{1}{16} - \frac{1}{24} \right) x$$

$$1500 = \frac{1}{48}x$$

$$x= 48 \cdot 1500 = 72000$$

--Elucidus