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A certain brand of car tyres last 450 km on a front wheel or 600 km on a rear wheel. By swapping the front and rear tyres, what is the largest distance in km that can be travelled using this set of four tyres?

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closed as off-topic by kingW3, Namaste, Arnaldo, Daniel W. Farlow, Shailesh May 6 '17 at 0:03

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  • $\begingroup$ Its unclear what happens when you switch the tires, assuming you swich it at 450km are you left with 450km,600km or 150km,0km? $\endgroup$ – kingW3 May 5 '17 at 11:36
  • $\begingroup$ If you switch at $450$ the fronts would be expired or $0%$ left. $\endgroup$ – marshal craft May 5 '17 at 11:41
  • $\begingroup$ Have a look at this for a similar question and how to solve math.stackexchange.com/questions/1762538/… $\endgroup$ – David Quinn May 5 '17 at 12:27
  • $\begingroup$ Or can the rears go deeper into the tire? $\endgroup$ – marshal craft May 5 '17 at 13:30
  • $\begingroup$ @DavidQuinn +1 for viewing the problem in terms of tires-per-mile. $\endgroup$ – amd May 5 '17 at 18:25
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The car can certainly go at least $450$ and no more than $600$ km on one set of tires, regardless of how the tires are swapped, so the function $r=f(S)$ that gives the range $r$ of a swapping strategy $S$ is bounded and has a maximum value over all swapping strategies. Call this optimal range $K$. ($f$ is continuous in the sense that a tiny modification of the swapping strategy will have a tiny effect on the range.)

All together, the tires spend a total $4K$ tire-km on the road. The total number of “front-km” equals the number of “rear-km,” so $2K$ of these km are front-km, and the average number of front-km a tire spends on the road is $K/2$. Some tire must therefore spend at least $K/2$ km on the front. If any tire spends more than $K/2$ miles on the front, another tire spends less than $K/2$ on the front and is not fully worn at the end of the trip (because it had a less weary front-to-back ratio), so the swapping strategy can be improved if these two tires are swapped to equalize their front and back km.

This argument implies that $K$ is attained when each tire accumulates exactly $K/2$ front-km (and consequently $K-K/2=K/2$ back-km). A tire wears down $1/450$ of its tread for each front km and $1/600$ of its tread for each back km. Therefore each tire wears down $\frac{K}2/450 + \frac{K}2/600$ of its tread in all. The tires wear down completely after $K$ km, so $\frac{K}2/450 + \frac{K}2/600=1$, whence $K=3600/7$.

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  • $\begingroup$ Swapping front with back once after half the distance would do it. $\endgroup$ – Steve Kass May 5 '17 at 18:02
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Ok if I interpret the question so that it is interesting (as currently phrased it is trivial that answer is, it doesnt matter the max distance is 450) then we ask if the front wears through 100% of the tire in 450 km and back in 600 km then what is the maximum distance one can travel on good tires provided they swap the tires exactly one time?

We now consider the front and back separately for swap $x\le 450$,

$$D_{max \ f}(x)= 450\frac{600-x}{600}+x$$

$$D_{max \ r}(x)=600\frac{450-x}{450}+x$$

Now for final max we take $$D_{max}(x)=MIN(D_{max \ f}(x), D_{max \ r}(x))$$

The question now is to find the maximum of $D_{max}(x)$

For $x \le 300$, $D_{max}(x)=D_{max \ f}(x)$ and for $x \ge 300$, $D_{max}(x)=D_{max \ r}(x)$.

We can see a clear maximum at $x=300$ which would be the best place to swap yielding $500 \ km$ total. enter image description here

Edit I've made a bit of a clerical error in the algebra but the procedure is sound.

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  • $\begingroup$ *image is from wolframalpha.com and subject to all of there regulations on it. $\endgroup$ – marshal craft May 5 '17 at 16:32

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