Choosing Between Three Shuttle-Van Quotes
Southview Middle School is renting shuttle vans for a field trip to a science museum. Company A's quote lists only three data points: 40 miles costs $118, 70 miles costs $139, and 100 miles costs $160. Company B sends an equation instead: C = 55 + 1.05m, where C is the total charge in dollars and m is the one-way mileage. A third option, Company C, charges a flat $200 for any one-way trip of 250 miles or less, with no per-mile charge at all. The trip coordinator has budgeted $210 each way and wants the least expensive option for the 180-mile trip, but she also needs a cheaper choice for a separate 60-mile supply run. A parent has told her flatly that “Company B can never be the cheaper choice,” and she wants that claim checked rather than trusted.
1. Use Company A's data to write a linear rule for C in terms of m and explain what makes the table linear. Then determine the cheapest option for the 60-mile supply run.
Read the worked answer
Check the rate of change between consecutive points.
- From 40 to 70 miles: (139 − 118) ÷ (70 − 40) = 21 ÷ 30 = $0.70 per mile.
- From 70 to 100 miles: (160 − 139) ÷ (100 − 70) = 21 ÷ 30 = $0.70 per mile.
A constant rate of change of $0.70 per mile means the relationship is linear. Using point-slope form with (40, 118): C = 0.70(m − 40) + 118 = 0.70m + 90. Check at m = 100: 0.70(100) + 90 = $160, which matches the table.
At 60 miles:
- Company A: 0.70(60) + 90 = 42 + 90 = $132.
- Company B: 55 + 1.05(60) = 55 + 63 = $118.
- Company C: $200 flat.
Company B is cheapest for the 60-mile run, $14 below Company A.
2. Solve the system to find the mileage where Companies A and B charge the same amount. Then compare all three companies for the 180-mile trip and evaluate the parent's claim that Company B can never be cheaper.
Read the worked answer
Set the rules equal: 0.70m + 90 = 1.05m + 55.
Subtract 0.70m from both sides: 90 = 0.35m + 55. Subtract 55: 35 = 0.35m. Divide: m = 100 miles.
At exactly 100 miles both charge 0.70(100) + 90 = $160, consistent with the table.
Compare the two slopes and starting values: Company B has the lower fixed charge ($55 versus $90) but the higher per-mile rate ($1.05 versus $0.70). Below 100 miles the fixed charge dominates and B is cheaper; above 100 miles the mileage rate dominates and A is cheaper. The parent's claim is therefore false - B wins for any trip under 100 miles.
For 180 miles:
- Company A: 0.70(180) + 90 = 126 + 90 = $216.
- Company B: 55 + 1.05(180) = 55 + 189 = $244.
- Company C: $200 flat.
Company C is cheapest and is the only option under the $210 budget. Reasoning from the two linear rules alone would have picked A and blown the budget by $6.
3. Is total charge a function of mileage for each company? Explain, and interpret the $90 in Company A's rule in context.
Read the worked answer
Yes for all three. In each plan, a given mileage produces exactly one total charge, so each relationship passes the definition of a function. Companies A and B are linear functions with different rates and starting values; Company C is a constant function with slope 0 on the domain 0 < m ≤ 250, and it stops being a valid model once a trip exceeds 250 miles - a reminder that a function is only meaningful on its stated domain.
The $90 is Company A's fixed charge: the amount owed before any miles are driven. It is the y-intercept of the rule and it is what makes A expensive for short trips.
One caution: the quoted data begin at 40 miles. Substituting m = 0 returns $90, but that is an extrapolation of the pattern, not a documented price for a zero-mile trip. If a booking fee or minimum charge changes below 40 miles, the linear model would mislead.