The Jeep Problem | FractionRush

🚙 The Setup

Topics: optimisation, harmonic series, fractions, logical reasoning

A jeep must cross a desert. It has a fuel tank that holds exactly 1 unit of fuel, and it travels exactly 1 unit of distance per unit of fuel.

There is an unlimited supply of fuel at the starting point (the edge of the desert), but there is no fuel anywhere else — unless the jeep carries it there first and caches it.

The jeep can make as many trips as needed to ferry fuel into the desert.

Question: Using n full tanks of fuel total, what is the maximum distance the jeep can travel into the desert (one-way)?

Optimal strategy: Work backwards from n tanks. Divide the journey into segments.

With n cans total, the maximum distance = 1/2 × (1 + 1/3 + 1/5 + … + 1/(2n−1))

This is half the sum of the first n odd reciprocals. As n → ∞, this sum grows without bound — meaning with enough fuel, the jeep can travel any distance!

With 2 full tanks:
Trip 1: Drive 1/4 forward, cache 1/2 tank there, drive 1/4 back. Used: 3/4. Cache holds: 1/2.
Trip 2: Fill up at start (full tank = 1). Drive to cache (1/4), refuel to full (3/4 in tank + 1/2 cache → 1 full tank). Drive 3/4 further.
Total distance = 1/4 + 3/4 = 1 unit.

Formula check: 1/2 × (1 + 1/3) = 1/2 × 4/3 = 2/3... wait — the simpler version: with 2 tanks, max = 1/2 + 1/6 = 2/3 (using optimal fractional caching). This is a simplified model.

📝 Questions

With just 1 full tank of fuel (no caching possible), what is the maximum distance the jeep can travel into the desert? 1 mark

With 1 tank, why can't the jeep cache fuel and come back? What constraint prevents this? 2 marks

The maximum distance with n tanks is: d(n) = ½(1 + ⅓ + ⅕ + … + 1/(2n−1)). Calculate d(1), d(2), d(3), and d(4). Leave answers as fractions. 4 marks

n (tanks)	1	2	3	4
d(n)

As n increases, does d(n) approach a maximum limit? Or does it keep growing? Describe the behaviour. 3 marks

Approximately how many tanks would be needed to travel 2 units into the desert? Use trial and error with your formula. 4 marks

The harmonic series 1 + ½ + ⅓ + ¼ + … also diverges (grows without bound). Explain the connection between the jeep problem and the harmonic series. 4 marks

Extension: In a simpler version: the jeep can only make whole trips (not partial). With 3 full tanks, what is the maximum integer number of units it can travel? Draw a diagram of your fuel caching plan. 4 marks

Answer Key

🚙 Jeep Problem

1.With 1 tank, maximum distance = 1 unit (drive straight across, no return needed).

2.To cache and return, the jeep would use fuel for the round trip to the cache point. With exactly 1 tank, any fuel used going forward and returning leaves less than a full tank for the final push — so there's no net gain unless starting with more fuel.

3.d(1) = ½ × 1 = 1/2. d(2) = ½(1+⅓) = ½×4/3 = 2/3. d(3) = ½(1+⅓+⅕) = ½×23/15 = 23/30. d(4) = ½(1+⅓+⅕+1/7) = ½×176/105 = 88/105.

4.d(n) keeps growing without bound — there is no maximum limit. The series ½(1+⅓+⅕+…) diverges, so given enough fuel, the jeep can reach any distance.

5.d(n) = 2 requires ½ × Σ(1/(2k−1)) = 2 → Σ = 4. The sum 1+⅓+⅕+1/7+1/9+… reaches 4 around n ≈ 15,800 tanks.

6.The jeep formula involves odd reciprocals; the harmonic series involves all reciprocals. Both diverge. The connection: in both cases, adding more terms (more tanks/more terms) always increases the sum, but ever more slowly — yet there is no ceiling.

7.With 3 whole tanks, a practical plan: Trip 1+2 ferry fuel to a cache 1/3 unit in. Trip 3 uses the cache to extend range. Maximum practical distance ≈ 1 + 1/3 + 1/5 (using the formula's insight) ≈ 1.53 units if fractional caching is allowed; roughly 1 unit with only whole trips. Accept reasonable diagrams.

🚙 The Jeep Problem

Answer Key

🚙 Jeep Problem

Answer Key

Jeep Problem