Ride-Hailing Tariffs with a Long-Distance Surcharge
Lina commutes between home and a study centre, a one-way distance of x km, where 0 < x ≤ 20. Two ride-hailing apps quote different tariffs. App A charges a fixed booking fee of 2.50 units plus 0.80 units per kilometre. App B charges a fixed fee of 4.00 units plus 0.60 units per kilometre, but for any trip longer than 10 km it adds a 1.20-unit long-distance surcharge to the whole fare. Lina needs a function model for each app valid across the whole range, and she has a daily budget of 18 units. She notices that at exactly 10 km the two apps are not equally priced and wonders whether the cheaper app can change more than once. She is preparing a short report and must justify her choice with inequalities rather than by testing a few distances. Her report must also state the maximum distance she can travel on 18 units under each app, and whether insisting on the cheaper app changes that maximum.
1. Write C_A(x) and C_B(x) as piecewise functions on 0 < x ≤ 20, and solve C_A(x) = C_B(x).
Read the worked answer
App A has no break, so
[C_A(x)=2.50+0.80x.]
App B changes at 10 km:
[C_B(x)=\begin{cases}4.00+0.60x, & 0<x\le 10,\ 4.00+0.60x+1.20=5.20+0.60x, & 10<x\le 20.\end{cases}]
For 0 < x ≤ 10: [2.50+0.80x=4.00+0.60x \Rightarrow 0.20x=1.50 \Rightarrow x=7.5.] This is valid.
For 10 < x ≤ 20: [2.50+0.80x=5.20+0.60x \Rightarrow 0.20x=2.70 \Rightarrow x=13.5.] This is also valid.
At x = 10 the fares are not equal: (C_A(10)=10.50) and (C_B(10)=10.00). So the equality points are x = 7.5 and x = 13.5, with a jump at x = 10.
2. Solve the inequalities C_A(x) < C_B(x) and C_B(x) < C_A(x). Interpret the result, including the point x = 10.
Read the worked answer
On 0 < x ≤ 10, (C_A<C_B) means [2.50+0.80x<4.00+0.60x \Rightarrow x<7.5.] Thus App A is cheaper for (0<x<7.5), and App B is cheaper for (7.5<x\le10). At x = 7.5 they are equal.
On 10 < x ≤ 20, (C_A<C_B) means [2.50+0.80x<5.20+0.60x \Rightarrow x<13.5.] Thus App A is cheaper again for (10<x<13.5), while App B is cheaper for (13.5<x\le20). At x = 13.5 they are equal.
Combining:
- App A cheaper: ((0,7.5)\cup(10,13.5));
- App B cheaper: ((7.5,10]\cup(13.5,20]);
- equal: (x=7.5) and (x=13.5).
The surcharge creates a jump at x = 10, so the cheaper app changes more than once. Testing only a few distances could easily miss the second switch.
3. Find the maximum distance Lina can travel on 18 units using App A, and then using the cheaper app at each distance. Does the cheaper app increase the maximum distance?
Read the worked answer
Using App A only: [2.50+0.80x\le18 \Rightarrow 0.80x\le15.50 \Rightarrow x\le19.375.] So App A allows at most 19.375 km on 18 units.
Using the cheaper app, check each interval:
- On ((0,7.5)), the maximum cost is at x = 7.5, (C_A=8.50), well under 18.
- On ((7.5,10]), the maximum cost is at x = 10, (C_B=10.00), under 18.
- On ((10,13.5)), the maximum cost is at x = 13.5, (C_A=13.30), under 18.
- On ((13.5,20]), (C_B(x)=5.20+0.60x). At x = 20, (C_B=17.20\le18).
Therefore the full stated range 0 < x ≤ 20 is affordable under the cheaper app, so the maximum distance is 20 km. The cheaper app raises the maximum from 19.375 km to 20 km, but only because App B remains under budget at the top end.